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THE UNIVERSITY 
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POOL ALGHBRA. 


UNIVERSTiy OF Iki FAQe LIBRAR) 
ANY 


G.: A. WENTWORTH, 


PROFESSOR OF MATHEMATICS IN PHILLIPS EXETER ACADEMY. 


BOSTON, U.S.A.: 
PUBLISHED BY GINN & COMPANY. 
LSode 








G. A. WENT WORTH, 


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in the Office of the Librarian of Congress, at Washington. 
VASA OE SS, 
A i} ALL RIGHTS RESERVED. 


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Typograpuy BY J. 8. Cusnine & Co., Boston, U_ 


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PREFACE. 


SSS 


Tus book, as the name implies, is written for High Schools and 
Academies, and is a thorough and practical treatment of the prin-. 
ciples of Elementary Algebra. It covers sufficient ground for 
admission to any American college, and with the author’s Col- 
lege Algebra makes as extended a course as the time allotted 
to this study in our best schools and colleges will allow. Great 
care has been taken to present the best methods, so that students 
in going from the lower book to the higher will have a good 
foundation, and have nothing to unlearn. 

The problems are carefully graded. They are for the most part 
new; either original or selected from recent examination papers. 
They are sufficiently varied and interesting, and are not so difficult 
as to discourage the beginner. The early chapters are quite full; 
for even if a student is perfectly familiar with the operations of 
Arithmetic, he must have time to learn the language and the 
fundamental processes of Algebra. 

The introductory chapter should be read and discussed in the 
recitation room. This chapter brings before the student in brief 
review the knowledge he has already gained from the study of 
Arithmetic, states and proves the general laws of numbers, sets 
forth clearly the advantage of using letters to represent numbers 
in the statement of general laws, and leads him to see at the 
outset that Algebra, like Arithmetic, treats of numbers. In this 
chapter, also, the meaning of negative quantities is explained, and 
the laws which regulate the combinations of different arithmetical 


numbers are shown to apply to algebraic numbers. It is hoped 


6) p- par oe 


VILUDO 


1V PREFACE. 


that a free discussion of these elementary principles will do much 
to prevent that vagueness which the beginner invariably experi- 
ences if he fails to connect the laws of Algebra with what he has 
learned in Arithmetic. 

Answers to the problems are bound separately, in paper covers, 
and will be furnished free to pupils when teachers apply to the 
publishers for them. | 

Any corrections or suggestions relating to the work will be 
thankfully received. 

G. A. WENTWORTH. 
PHILLIPS EXETER ACADEMY, 
June, 1890. 


= 


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CHAPTER 


CONTENTS. 


INTRODUCTION . 

ADDITION AND SUBTRACTION . 
MULTIPLICATION 

DIVISION . 

SIMPLE EQUATIONS 

MULTIPLICATION AND DIVISION . 

FAcTORS . 

Common Factors AND MULTIPLES . 
FRACTIONS 

FRACTIONAL EQUATIONS 

SIMULTANEOUS EQUATIONS OF THE First DEGREE 
PROBLEMS INVOLVING Two UNKNowN NuMBERS . 
INEQUALITIES 

INVOLUTION AND EvVouurion . 

THEORY OF EXPONENTS 

RapicAL EXPRESSIONS . 

IMAGINARY EXPRESSIONS . 

QUADRATIC EQuATIONS . 

SIMULTANEOUS QUADRATIC EQUATIONS 
PROPERTIES OF QUADRATICS . 

RATIO, PROPORTION, AND VARIATION . 
PROGRESSIONS 

BREE CMOS OM PERTES x6 yarns . vet an ter 
BInoMIAL THEOREM . 

LoGARITHMS 

GENERAL REVIEW EXERCISE. 


105 
120 
144 
163 
179 
197 
200 
217 
225 
243 
249 
277 
287 
293 
310 
323 
330 
342 
356 


SCHOOL ALGHBRA. 


CHAPTER I. 
INTRODUCTION. 


1, Units. In counting separate objects the standards by 
which we count are called units; and in measuring contin- 
uous magnitudes the standards by which we measure are 
called units. Thus, in counting the boys in a school, the 
unit is a boy; in selling eggs by the dozen, the unit is a 
dozen eggs; in selling cloth by the yard, the unit is a yard 
of cloth; in measuring short distances, the unit is an inch, 
a foot, or a yard; in measuring long distances, the unit 1s 
a rod or a mile. 


2. Numbers. Repetitions of the unit are expressed by 
numbers. If a man, in sawing logs into boards, wishes to 
keep a count of the logs, he makes a straight mark for 
every log sawed, and his record at different times will be 
as follows: 


Ree TOI TY | 
Beppe ELD APT HE be TALI 


These representative groups are named one, two, three, 
four, five, six, seven, eight, nine, ten, etc., and are known 
collectively under the general name of numbers, It is 
obvious that these representative groups will have the 
same meaning, whatever the nature of the unit counted. 


2 SCHOOL ALGEBRA. 


3. Quantities. The word ‘quantity’ (from the Latin 
guantus, how much) implies both a unit and a number. 
Thus, if we inquire how much wheat a bin will hold, we 
mean how many bushels of wheat it will hold. If we 
inquire how much carpeting there is in a certain roll, 
we mean how many yards of carpeting. If we inquire 
how much wood there is on a certain wood-lot, we mean 
how many cords of wood. 


4, Number-Symbols in Arithmetic. Instead of groups of 
straight marks, we use in Arithmetic the arbitrary sym- 
bols 1, 2, 8, 4, 5, 6, 7, 8, 9, called figures, for the numbers 
one, two, three, four, five, six, seven, eight, nine. 

The next number, ten, is indicated by writing the figure 
1 in a different position, so that it shall signify not one, but 
ten. This change of position is effected by introducing a 
new symbol, 0, called nought or zero, and signifying none. 
Thus, in the symbol 10, the figure 1 occupying the second 
place from the right, signifies a collection of ten things, and 
the zero signifies that there are no single things over. The 
symbol 11 denotes a collection of ten things and one thing 
besides. All succeeding numbers up to the number con- 
sisting of 10 tens are expressed by writing the figure for 
the number of tens they contain in the second place from 
the right, and the figure for the number of units besides in 
the first place. The number consisting of 10 tens is called 
a hundred, and the hundreds of a number are written in the 
third place from the right. The number consisting of 10 
hundreds is called a thousand, and the thousands are writ- 
ten in the fowrth place from the right; and so on. 


5. The Natural Series of Numbers. Beginning with the 
number one, each succeeding number is obtained by put- 
ting one more with the preceding number. If from a given 


INTRODUCTION. + 


point marked 0, we draw a straight line to the right, and 
beginning from this point lay off units of length, the suc- 
cessive repetitions of the unit will be denoted by the natural 
series of numbers 1, 2, 3, 4, etc. Thus, 


1 2 a 4 5 6 te. 
ee | : 5 rf ete 





6. The reader will notice that number symbols in Arith- 
metic stand for particular numbers, and that these symbols 
indicate a method of making up the number, but not neces- 
sarily the method by which the number is actually made 
up. Thus, if a man has 66 dollars in bank-notes, he may 
have, as the number 66 indicates, 6 ten-dollar bills and 6 
one-dollar bills, but this is not the only way in which the 
66 dollars may be made up. 


7. Integral and Fractional Numbers. When the things 
counted are whole units, the numbers which count them 
are called whole numbers, integral numbers, or integers, where 
the adjective is transferred from the things counted to the 
numbers which count them. But if the things counted are 
only parts of units, the numbers which count them are 
called fractional numbers, or simply fractions, where again 
the adjective is transferred from the things counted to the 
numbers which count them. 

To represent the parts of a given unit, two number- 
symbols are used, one to name the parts into which the 
unit is divided, and therefore called the denominator, and 
the other to denote the number of parts taken, and there- 
fore called the numerator. The denominator is written 
below the numerator with a line between them. ‘Thus, in 
the fraction ~ the 9 shows that the unit is divided into 
nine equal parts, called ninths of the unit, and the 7 shows 
that seven of these equal parts are taken. 


4 SCHOOL ALGEBRA. 


8. Principal Signs of Operations. The sign +, read plus, 
indicates that the number after the sign is to be added to 
the number before the sign. Thus, 5+4 means that 4 
is to be added to 5. 

The sign —, read minus, indicates that the number after 
the sign is to be subtracted from the number before the sign. 
Thus, 8—4 means that 4 is to be subtracted from 8. 

The sign X, read tames or into, indicates that the number 
after the sign is to be multiplied by the number before the 
sign. Thus, 5-x 4 means that 4 is to be multiplied by 5. 

The sign +, read divided by, indicates that the number 
before the sign is to be diwded by the number after the 
sign. Thus, 8+4 means that 8 is to be divided by 4. 

The operation of division is also indicated by placing 
the dividend over the divisor with a line between them. 
Thus, £ means the same as 8-4. 


9. Signs of Relation. The sign =, read equals, or is equal 
to, when placed between two numbers, indicates that they are 
equal. Thus, 8+4—12 means that 8+4 is the same as 12. 

The sign >, read is greater than, indicates that the num- 
ber which precedes the sign is greater than the number 
which follows it. Thus, 8+4-> 10 means that 8+ 4 is 
greater than 10. 

The sign <, read zs less than, indicates that the number 
which precedes the sign is less than the number which fol- 
lows it. Thus, 8+4<16 means that 8+4 is less than 16. 


10. Signs of Deduction and of Continuation. The sign .. 
stands for the word “therefore” or “hence.” The sign 
stone or -———-— stands for the words ‘and so on.” 


11. Number-Symbols in Algebra. Algebra, like Arith- 
metic, treats of numbers, and employs the letters of the 
alphabet in addition to the figures of Arithmetic to represent 


INTRODUCTION. 5 


numbers. The letters of the alphabet are used as general 
symbols of numbers to which any particular values may be 
assigned. In any particular problem, however, a letter must 
be supposed to have the same particular value throughout 
the investigation or discussion of the problem. 

These general symbols are of great advantage in investi- 
gating and stating general laws; in exhibiting the actual 
method in which a number is made up; and in represent- 
ing unknown numbers which are to be discovered from their 
relations to known numbers. 

. The advantage of representing numbers by letters will 
be more clearly seen later on. For the present it will be 
sufficient for the beginner to understand that every letter, 
and every combination of letters, and every combination of 
figures and letters used in Algebra, represents some number. 
Thus, the number of dollars in a package of bank-notes 
can be represented by x; but if the package consists of ten- 
dollar bills, five-dollar bills, two-dollar bills, and one-dollar 
bills, and if we denote the number of ten-dollar bills by a, 
of five-dollar bills by 0, of two-dollar bills by ¢, and of one- 
dollar bills by d, the whole number of dollars in the package 
will be represented by 10a+ 56+ 2c+d. 

In this particular case and 10a+56+2ec+d both 


stand for the same number. 


12. Substitution. It is obvious that the same operation 
on each of the above expressions will produce results that 
agree in value, and therefore that either may be substituted 
for the other at pleasure. In short, 

Every algebraic expression represents some number, and 
may be operated upon as uf it were a single symbol standing 
for the number which tt represents. 


18, Factors. When a number consists of the product of two 
or more numbers, each of these numbers is called a factor of 


6 SCHOOL ALGEBRA. 


the product. If these numbers are denoted by letters, the 
sign X is omitted. Thus, instead of a x b, we write ad. 


14, Coefficients. A known factor prefixed to another 
factor to show how many times that factor is taken is called 
a coefficient. 


15. Powers. A product consisting of two or more equal 
factors is called a power of that factor. 

The index or exponent of a power is a small figure placed 
at the right of a number, to show how many times the 
number is taken as a factor. Thus, 2‘ is written instead of 
2xX2x2~x 2; a instead of aaa. 

The second power of a number is generally called the 
square of that number; the third power of a number, the 
cube of that number. 


16. Roots. The root of a number is one of the equal fac- 
tors of that number; the sguare root of a number is one 
of the éwo equal factors of that number; the cube voot of a 
number is one of the three equal factors of that number ; 
and so on. The sign \/, called the radical sign, indicates 
that a root is to be found. Thus, 4, or V4, means that 
the square root of 4 is to be taken; 8 means that the 
cube root of 8 is to be taken; and so on. 

The figure written above the radical sign is called the 
index of the root. 


17. An algebraic expression is a number written with alge- 
braic symbols; an algebraic expression consists of one sym- 
bol, or of several symbols connected by signs of operation. 

A term is an algebraic expression the parts of which are 
not separated by the sign of addition or subtraction. Thus, 
3ab, 52x 4y, 8ab+4 zy are terms. 


INTRODUCTION. 3 


A simple expression is an expression of one term. 
A compound expression is an expression of two or more 
terms. 


18. Positive and Negative Terms. The terms of a com- 
pound expression preceded by the sign + are called posi- 
tive terms, and the terms preceded by the sign — are called 
negative terms. The sign ++ before the first term is omitted. 


19, Parentheses. If a compound expression is to be 
treated as a whole it is enclosed in a parenthesis. Thus, 
2x (10+ 5) means that we are to add 5 to 10 and multiply 
the result by 2; if we were to omit the parenthesis and 
write 2x 10+ 5, the meaning would be that we were to 
multiply 10 by 2 and add 5 to the result. 

Instead of parentheses, we use with the same meaning 
brackets [ ], braces {?, and a straight line called a vinculum. | 


Thus, (5+ 2), [5+ 2], {5+ 23, 5+2, ave) all mean that 
the expression 5-+ 2 is to be treated as the single symbol 7. 


20. Rules for removing Parentheses. If a man has 10 dol- 
lars and afterwards collects 8 dollars and then 2 dollars, 
it makes no difference whether he adds the 8 dollars to his 
10 dollars, and then the 2 dollars, or puts the 8 and 2 
dollars together and adds their sum to his 10 dollars. 

The first process is represented by 10+ 3+ 2. 

The second process is represented by 10-+(8 + 2). 

Hence 10+(8+ 2)=10+ 3+ 2. (1) 

If a man has 10 dollars and afterwards collects 3 dol- 
lars and then pays a bill of 2 dollars, it makes no differ- 
ence whether he adds the 8 dollars collected to his 10 
dollars and pays out of this sum his bill of 2 dollars, or 
pays the 2 dollars from the 3 dollars collected and adds 
the remainder to his 10 dollars. 


8 SCHOOL ALGEBRA. 


The first process 1s represented by 10+ 3 — 2. 
The second process is represented by 10 + (8 — 2). 


Hence 10+(38—2)=10+3—2. (2) 


From (1) and (2) it follows that if a compound expres- 
sion is to be added, the parenthesis may be removed and 
each term in the parenthesis retain its prefixed sign. 

If a man has 10 dollars and has to pay two bills, one of 
3 dollars and one of 2 dollars, it makes no difference whether — 
he takes 8 dollars and 2 dollars in succession, or takes the 3 
and 2 dollars at one time, from his 10 dollars. 

The first process is represented by 10—3— 2. 

The second process is represented by 10 —(8+ 2). 


Hence 10—(8+ 2)=10—38 — 2. (3) 


If a man has 10 dollars consisting of 2 five-dollar bills, 
and has a debt of 3 dollars to pay, he can pay his debt by 
giving a five-dollar bill and receiving 2 dollars. 

This process is represented by 10 — 5 + 2. 

Since the debt paid is three dollars, that is, (6—2) dol- 
lars, the number of dollars he has left can evidently be 


expressed by 10 —(5~2) 
Hence 10 — (5 — 2) = 10—5+4 2. (4) 


From (8) and (4) it follows that if a compound expres- 
sion is to be subtracted, the parenthesis may be removed, 
provided the sign before each term within the parenthesis 
is changed, the sign + to —, and the sign — to +. 


Exercise 1. 
Perform the operations indicated, and simplify . 
1 7+(8—2). 4. 5x(2+8). 7. (7—38)x (5—2). 
2. 7T—(8--2). 5. (54+8)+2. 8. (8—2)+(5—2). 
38. 7—(8+2). 6 5x(38—2). 9. 3xX(2—6—2). 


INTRODUCTION. 9 


21, Fundamental Laws of Numbers. We are so occupied 
in Arithmetic with the application of numbers to the ordi- 
nary problems of every-day life that we pay little attention 
to the investigation of the fundamental laws of numbers. 
It is, however, very important that the beginner in Algebra 
should have clear ideas of these laws, and of the extended 
meaning which it is necessary to give in Algebra to cer- 
tain words and signs used in Arithmetic; and that he 
should see that every such extension of meaning is con- | 
sistent with the meaning previously attached to the word 
or sign, and with the general laws of numbers. We shall, 
therefore, give general definitions for the fundamental oper- 
ations upon numbers and then state the laws which apply 
to them. 


22, Addition. ‘The process of finding the result when 
two or more numbers are taken together is called addition, 
and the result is called the sum. 


28, Subtraction. The process of finding the result when 

one number is taken from another is called subtraction, and 
the result is called the difference or remainder. ‘The number 
taken away is called the subtrahend, and the number from 
which the subtrahend is taken is called the minuend. 

In practice the difference is found by discovering the 
number which must be added to the subtrahend to give the 
minuend. Ifthe subtrahend consists of two or more terms, - 
we add these terms and then determine the number which 
must be added to their sum to make it equal to the minu- 
end. Thus, if a clerk in a store sells articles for 10 cents, 
15 cents, and 30 cents, and receives a dollar bill in pay- 
ment, he makes change by adding these items and then 
adding to their sum enough change to make a dollar. 

From the nature of this process it is obvious that the 
general laws of numbers which apply to addition apply 


10 SCHOOL ALGEBRA. 


also to subtraction, and that we may take for the general 
definition of subtraction 


The operation of finding from two given numbers, called 
minuend and subtrahend, a third number, called difference, 
which added to the subtrahend will give the minuend. 


24. Multiplication. The process of finding the result 
when a given number is taken as many times as there are 
units in another number is called multiplication, and the re- 
sult is called the product. 

This definition fails when the multiplier is a fraction, for 
we cannot take the multiplicand a fraction of a tume. We 
therefore consider what extension of the meaning of multi- 
plication can be made so as to cover the case in question. 
When we multiply by a fraction we divide the multiplicand 
into as many equal parts as there are units in the denomi- 
nator and take as many of these parts as there are units 
in the numerator. If, for instance, we multiply 8 by #, we 
divide 8 into four equal parts and take three of these parts, 
getting 6 for the product. We see that ¢ is ¢ of 1, and 6 
is $ of 8; that is, the product 6 is obtained from the mul- 
tiplicand 8 precisely as the multiplier 2 is obtained from 1. 
The same is true when the multiplier is an integral number. 


Thus, in 6 X 8= 48, 
the multiplier 618 1-14 1-4-1 ae 
and the product 48 is 8+8+8+48+48+8. 


We may, therefore, take for the general definition of 
multiplication 


The operation of finding from two given numbers, called 
multiplicand and multepler, a third number called product, 
which is formed from the multiplicand as the multipher is 
formed from unity. 


INTRODUCTION. ‘ 11 


25. Division. To divide 48 by 8 is to find the number 
of times it 1s necessary to take 8 to make 48. Here the 
product and one factor are given and the other factor is 
required. We may therefore take for the general definition 
of division 

The operation by which when the product and one factor 
are given the other factor is found. 


With reference to this operation the product is called 
the dividend, the given factor the divisor, and the required: 
factor the quotient. 


26. The Commutative Law. If we have a group of 3 
things and another group of 4 things, we shall have a 
group of 7 things, whether we put the 3 things with the 
4 things or the 4 things with the 3 things; that is, 

4+38=344. 


It is evident that the truth of the above statement does 
not depend upon the particular numbers 3 and 4, but that 
the statement is true for any two numbers whatever. Thus, 
in case of any two numbers we shall have 


First number + second number = second number + first 
number. 


If we let a stand for the first number and 6 for the second 
number, this statement may be written in the much shorter 
form 

atb=b-+a. 

This is the commutative law of addition, and may be 
stated as follows: 

Additions may be performed in any order. 


27. Also, if we have 5 lines of dots with 10 dots in a 
line, the whole number of dots will be expressed by 5 x 10. 


12 ; SCHOOL ALGEBRA. 


If we consider the dots as 10 columns with 5 dots in a 
column, the number will be expressed by 10 x 5. 


That 1s, DLO Sh eco 
Again, if we divide a given length into 6 equal parts, 


3 


eel 


1 
2 








one-third of the line will contain 2 of these parts, and one- 
half the line will contain 8 of these parts. Now one-third 
of one-half will be 1 of these parts, and one-half of one- 
third will be 1 of these parts; that is, 


tXt=tXF. 
Therefore, if a and 6 stand for any two numbers, integral 
or fractional, we shall have 
ab = ba. 
This is the commutative law of multiplication, and may 
be stated as follows: 


Multipheations may be performed in any order. 


28. The Distributive Law. The expression 4 x (5+ 38) 
means that we are to take the sum of the numbers 5 and 3 
four times. ‘The process can be represented by placing five 
dots in a line, and a little to the right three more dots in 
the same line, and then placing a second, third, and fourth 
line of dots underneath the first line and exactly similar to 


it, eocee50 


° 


INTRODUCTION. 13 


There are (5-++ 3) dots in each line, and 4 lines. The 
total number of dots, therefore, is 4 x (5 + 8). 

We see that in the left-hand group there are 4 x 5 dots, 
and in the right-hand group 4 x 3 dots. The sum of these 
two numbers (4 X 5)-+ (4 x 3) must be equal to the total 
number; that is, 


4x (5+3)=(4x 5)+(4x 8). 


Again, the expression 4 x (8 — 3) means that 3 is to be 
taken from 8, and the remainder ‘to be multiplied by 4. 
The process can be represented by placing eight dots in a 
line and crossing the last three, and then placing a second, 
third, and fourth line of dots underneath the first line and 
exactly similar to it. 


e@ee8 @ 
eee @ 
eee ®@ 
e@e38 @ 
e@®ee @ 
ew ea ea eB 
ee Oe ee OB 
ew ee 8 OB 


The eile. number of dots not crossed in each line is 
evidently (8 —3), and the whole number of lines is 4. 
Therefore the total number of dots not crossed is 

4x (8—8). 


The total number of dots (crossed and not crossed)-is 
(4 x 8), and the total number of dots crossed is (4 x 8). 
Therefore the total number of dots not crossed is 


Eee AO) (a8) ; 
that is, 4x (8—3)=(4 x 8)—(4 x 3). 
Hence, by the commutative law 
(8 —3)x4=(8 x 4)—(8 x 4). 
In like manner, if a, 6, ¢, and d stand for any numbers, 


we have ax(b+e—d)=ab+ace—ad, 


14 SCHOOL ALGEBRA. 


This is the distributive law, and may be stated as follows: 

In multiplying a compound expression by a simple ex- 
pression the result is obtained by multiplying each term of 
the compound expression by the simple expression, and writ- 
mg down the successwe products with the same signs as 
those of the original terms. 


29. The Associative Law. The terms of an expression may 
be grouped in any manner. For if we have several num- 
bers to be added, the result will evidently be the same, 
whether we add the numbers in succession or arrange them 
in groups and add the sums of these groups. Thus, 


at+btetdte 
=a-+(b+¢)+(d +e) 
=(a+8)+(et+dto) 

Likewise, if in the rectangular solid represented in the 
margin we suppose AB to contain 5 units of length, BC 3 
units, and CD 7 units. The base may 
be divided into square units. There 
will be 3 rows of 5 square units each. 
Upon each square unit a cubic unit may 
be formed, and we shall have (8x5) 
cubic units. Upon these another tier of 
(3 x 5) cubic units may be formed, and 
then another tier of the same number, 
and the process continued until we have 7 tiers of (8 X 5) 
cubic units. Hence the number of cubic units in the solid 
will be represented by 7 x (8 x 5). 

Upon the right-hand square in the back row a pile of 7 
cubic units may be formed, upon the next square to the 
left another pile of 7 cubic units may be formed, and 
upon the next square another, and the process continued 
until we have a pile of 7 cubic units on each square in the 


NENENE NENT EEN] 





INTRODUCTION. 15 


back row. We shall then have (5 x 7) cubic units in the back 
tier, and as we can have 8 such tiers, the number of: cubic 
units in the solid will be represented by 3 x (5 x 7). 
Again, if we form a pile of 7 cubic units on the right-hand 
square of the back row, then another pile of 7 cubic units on 
the next square in front, another pile of 7 cubic units on the 
next square in front, we shall have a tier of (3 x 7) cubic 
units. We can have 5 such tiers, and the number of cubic 
units in the solid will now be represented by 5 x (8 X 7). 
It follows, therefore, that the total number of cubic units 
in the solid may be represented by 
7X(8x 5), or by 3X(5~x 7), or by 5X (8X 7). 
It is obvious that no part of this proof depends upon the 
particular numbers 3, 5, and 7, but the law holds for any 
arithmetical numbers whatever, and may be expressed by 
Oxiax b)—aXx (bx ¢c)=5 X(aXc). 
This is called the associative law of addition and multi- 
plication, and may be stated as follows: 


The terms of an expression, or the factors of a product, 
may be grouped nm any manner. 


30. The Index Law. 
Since a =aa, and a’ = aaa, 
oer ax a a a 
6 NENT CEG TT cai | ee | haan 
If a stands for any number, and m and 2 for any integers, 
since a”= aaa: to m factors, 
and a” = aaa: to factors, 
a” X a" = (aaa ---- to m factors) X (aaa..... to n factors), 


= aaa: to (m+n) factors, 


16 . SCHOOL ALGEBRA. 


Hence, the index law may be stated as follows: 


The index of the product of two powers of the same number 
as equal to the sum of the indices of the factors. 


31. These four laws, the commutative, the distributive, 
the associative, and the index laws, are the fundamental 
laws of Arithmetic, and together with the daw of signs, 
which will be explained hereafter, they constitute the 
fundamental laws of Algebra. 


32. Quantities Opposite in Kind. If a man gains 6 dollars 
and then loses 4 dollars, his actual gain, or, as we com- 
monly say, his net gain, is 2 dollars; that is, 4 dollars’ loss 
cancels 4 dollars of the 6 dollars’ gain and leaves 2 dollars’ 
gain. If he gained 6 dollars and then lost 6 dollars, the 6 
dollars’ doss cancels the 6 dollars’ gain, and his net gaz is 
nothing. If he gained 6 dollars and then lost 9 dollars, 
the 6 dollars’ gain cancels 6 dollars of the 9 dollars’ loss, 
and his net loss is 8 dollars. In other words, loss and gain 
are quantities so related that one cancels the other wholly 
or in part. 

If the mercury in a thermometer rises 12 degrees and 
then falls 7 degrees, the fall of 7 degrees cancels 7 degrees 
of the vise, and the net rise is 5 degrees. If it rises 12 de- 
grees and then fal/s 12 degrees, the net rise is nothing. 
If it vuses 12 degrees and falls 15 degrees, there is a net 
fall of 3 degrees. In other words, rzse and fall are quan- 
tities so related that one cancels the other wholly or in 
part. 

An opposition of this kind also exists in motion forwards 
and motion backwards; in distances measured east and 
distances measured wes‘; in distances measured north and 
distances measured south; in assets and debts; in time be- 
fore and time after a fixed date; and so on. | 


INTRODUCTION. Ly 


33. Algebraic Numbers. If we wish to add 4 to 3, we 
begin at 4 in the natural series of numbers, 


Dette 804. B68 8 





count 3 units forwards, and arrive at 7,the sum sought. If 
we wish to subtract 3 from 7, we begin at 7 in the natural 
series of numbers, count 3 units backwards, and arrive at 4, 
the difference sought. If we wish to subtract 7 from 7, we 
begin at 7, count 7 units backwards, and arrive at 0. If 
we wish to subtract 7 from 4, we cannot do it, because 
when we have counted backwards as far as 0 the natural 
series of numbers comes to an end. 

In order to subtract a greater number from a smaller it 
is necessary to asswme a new series of numbers, beginning 
at zero and extending to the left of zero. The series to the 
left of zero must proceed from zero by the repetitions of the 
unit, precisely hke the natural series to the right of zero; 
and the opposition between the right-hand series and the 
left-hand series must be clearly marked. This opposition 
is indicated by calling every number in the right-hand 
series a positive number, and prefixing to it, when written, 
the sign +; and by calling every number in the left-hand 
series a negative number, and prefixing to it the sign —. 
The two series of numbers will be written thus: 


ae rem ee er PO Sap a8 4 Ae 


ALGEBRAIC SERIES OF NUMBERS. 


If, now, we wish to subtract 9 from 6, we begin at 6 in 
the positive series, count 9 units in the negative direction 
(to the left), and arrive at —3 in the negative series; that 
is, 6—9=— 8. 

The result obtained by subtracting a greater number from 
a less, when both are positive, is always a negatwe number, 


+ 


18 SCHOOL ALGEBRA. 


In general, if a and } represent any two numbers of the 
positive series, the expression a—6 will be a positive num- 
ber when a is greater than 6; will be zero when a is equal 
to 6; will be a negative number when a is less than 0. 

In counting from left to right in the algebraic series num- 
bers werease in magnitude; in counting from right to left 
numbers decrease in magnitude. Thus —38, —1, 0, +2, 
+4 are arranged in ascending order of magnitude. 


34. We may illustrate the use of algebraic numbers as 





follows $ =Se 0 8 90 
Spee Ee a= pes 
D A C 


Suppose a person starting at A walks 20 feet to the right 
of A, and then returns 12 feet, where will he be? Answer: 
At C,a point 8 feet to the right of A; that is, 20 feet —12 
feet = 8 feet; or, 20—12=8. 

Again, suppose he walks from A to the right 20 feet, and 
then returns 20 feet, where will he be? Answer: At A, 
the point from which he started; that is, 20 —20=0. 

Again, suppose he walks from A to the right 20 feet, and 
then returns 25 feet, where will he now be? Answer: At 
D, a point 5 feet to the left of A; that is, 20 —25=— 5; 
and the phrase ‘‘5 feet to the left of A’ is now expressed 
by the negative quantity, — 5 feet. 


35, Every algebraic number, as + 4 or —4, consists of a 
sign + or — and the absolute value of the number. The 
sign shows whether the number belongs to the positive or 
negative series of numbers; the absolute value shows what 
place the number has in the positive or negative series. 

When no sign stands before a number, the sign + is 
always understood. Thus 4 means the same as +4, a 
means the same as-+a. But the sign — is never omitted. 


INTRODUCTION. 19 


36. Two algebraic numbers which have, one the sign + 
and the other the sign —, are said to have unlike signs. 
. Two algebraic numbers which have the same absolute 
values, but unlike signs, always cancel each other when 


combined. Thus +4—4=0, +a—a=0. 


37. Double Meanings of the Signs + and —. The use of 
the signs + and — to indicate addition and subtraction 
must be carefully distinguished from the use of the signs ++ 
and — to indicate in which series, the positive or the nega- 
tive, a given number belongs. In the first sense they are 
signs af operations, and are common to Arithmetic and 
Algebra: in the second sense they are signs of opposition, 
atid are Beateeeds in Algebra alone. 


38, Addition and Subtraction of Algebraic Numbers. An 
algebraic number which is to be added or subtracted is 
often inclosed in a parenthesis, in order that the signs ++ 
and —, which are used to distinguish positive and negative 
numbers, may not be confounded with the + and — signs 
that denote the operations of addition and subtraction. 
Thus + 4 -+ (— 8) expresses the sum, and + 4 — (— 8) ex- 
presses the difference, of the numbers -++ 4 and — 8. 

In order to add two algebraic numbers we begin at the 
place in the series which the first number occupies and 
count, in the direction indicated by the sign of the second 
number, as many units as there are in the absolute value 
of the second number. 

Thus the sum of + 4-+ (+8) is found by counting from 
+4 three units in the positive direction; that is, to the 
right, and is, therefore, ++ 7. 

The sum of + 4+ (— 8) is found by counting from + 4 
three units in the negative direction; that is, to the left, and 


is, therefore, + 1. 


920 SCHOOL ALGEBRA. 


The sum of —4-+(4 8) is found by counting from — 4 


three units in the positive direction, and is, therefore, — 1. 


seeee aa ee a gems | 041442 +3 44 45 +6-— 


The sum of — 4-+ (— 3) is found by counting from — 4 
three units in the negative direction, and is, therefore, — 7. 

Hence to add two or more algebraic numbers, we have 
the following rules: 


Case I. When the numbers have lke signs. Find the 
sum of thewr absolute values, and prefix the common sign to 
the result. 


Case II. When there are two numbers with wnlcke signs. 
Find the difference of their absolute values, and prefix to the 
result the sign of the greater number. 


Case III. When there are more than two numbers with ~ 
unlike signs. Combine the first two numbers and thas result 
with the third number, and so on; or, find the sum of the 
positive numbers and the sum of the negative numbers, take 
the difference between the absolute values of these two sums, 
and prefix to the result the sign of the greater sum. 


39. The result is called the sum. It is often called the 
algebraic sum, to distinguish it from the arithmetical sum, 
that is, the sum of the absolute values of the numbers. 


40, Subtraction. In order to subtract one algebraic num- 
ber from another, we begin at the place in the series which 
the minuend occupies and count in the direction opposite to 
that wndicated by the sign of the subtrahend as many units 
as there are in the absolute value of the subtrahend. 

Thus, the result of subtracting + 3 from + 4 is found by 
counting from +4 three units in the negatwe durection; 
that is, in the direction opposite to that indicated by the sign 
+ before 8, and is, therefore, +1. 


INTRODUCTION. 21 


The result of subtracting —3 from +4 is found by count- 
ing from +4 three units in the positive direction; that is, 
in the direction opposite to that indicated by the sign — be- 
fore 3, and is, therefore, -+- 7. 

The result of subtracting +3 from —4 is found by count- 
ing from —-4 three units in the negative direction, and is, 
therefore, — 7. 

The result of subtracting —-3 from —4 is found by count- 
ing from —4 three units in the positive direction, and is, . 
therefore, — 1. 

Collecting the results obtained in addition and subtrac- 
tion, we have 


ADDITION. SUBTRACTION. 
+4-+-(—3)=+4—3=-+1. peeeeete eee} eereidemecr Ly 
Pers t448=4+7. +4. (—8)=44438=+47. 
pee 48> 7 4 (4 8)a + 4 827. 
eee ye 43 i, | 4 (8) 44 81 


No part of this proof depends upon the particular num- 
bers 4 and 3, and hence we may employ the general symbols 
a and 6 to represent the absolute values of any two ME 
braic numbers. We shall then have 


ADDITION. SUBTRACTION. 
+a+(—b)=+a—6. +-a—(+6)=+a—6. (1) 
+a+(+6)=+a-+0. +a—(—b)=+a+b6. (2) 
—a+(—b)=—a—b. —a—(+6)=—a—b. (8) 
—a+(+6)=—a+b. —a—(—b)=—a+b. (4) 


From (1) and (8), it is seen that subtracting a positwe 
wumber is equivalent to adding an equal negative number. 

From (2) and (4), it is seen that subtracting a negative 
number is equivalent to adding an equal positive number. 


pas SCHOOL ALGEBRA. 


To subtract one algebraic number from another, we have, 
therefore, the following rule : 


Change the sign of the subtrahend, and add the subtra- 
hend to the minuend. 


This rule is consistent with the definition of subtraction 
given in § 23; for, if we have to subtract — 4 from +3, we 
must add +4 to the subtrahend —4 to cancel it, and then 
add + 3 to obtain the minuend; that is, we must add +7 
to the subtrahend to get the minuend, but +7 is obtained 
by changing the sign of the subtrahend —4, making it +4, 
and adding it to +3, the minuend. 


41, The commutative law of addition applies to algebraic 
numbers, for +4-+(— 3)=—38-+(+4). In the first case 
we begin at +4 in the series, count three units to the left, 
and arrive at +1; in the second case we begin at —8 in 
the series, count four units to the right, and arrive at +1. 

The associative law, also, of addition is easily seen to 
apply to algebraic numbers. 


42. Multiplication and Division of Algebraic Numbers. By 
the definition of multiplication, § 24. 


Since +38=+1+4+1+1; 
“3X (+ 8)=4+8+4848 
=-+ 24, 
and 3x (—8)=—8—8—8 
= — 24, 
Again, since —s=—1—-1-l1; 
“.(—38)X 8=—8—8-—8 
= — 24, 
and = (= 8)x(—8)=—=(—8)—(- 8) -( 8) 
=+8+8+8 


Sao. 


INTRODUCTION. ; 23 


No part of this proof depends upon the particular num- 
bers 3 and 8. If we use a to represent the absolute value 
of any number, and 6 to represent the absolute value of 
any other number, we shall have 


(+a) x (+ 6)=+ ab. | (1) 
(+a) x (— 6) =— ab. (2) 
(=a) x (+ 6) = — ab. (3) 
(— a) X (—b)=+ ab. (4) » 


43, Law of Signs in Multiplication. From these four cases 
it follows that, in finding the product of two algebraic 
numbers, 


Lake signs give +, and unlwke signs gwe —. 


44, Law of Signs in Division. 


Since (+a) x(+6)=+ab, «. +ab+(+a)=+6. 
Since (+a) x (— 6) =—ab, «. —ab+(+a)=—8. 
Since (—a) X (+0)=—ab, .. —ab+(—a)=+0. 
Since (—a) x (—b)=+ ab, «. +ab-+(—a)=—0b. 


That is, if the dividend and divisor have like signs, the 
quotient has the sign-+; and if they have unlike signs, 
the quotient has. the sign —. Hence, in division, 


TInke signs gwe +; unlike signs gwe —. 


45. From the four cases of multiplication that we have 
given in § 42 it will be seen that the absolute value of 
each product is zndependent of the signs, and that the signs 
are independent of the order of the factors. Hence the com- 
mutative and associative laws of multiplication hold for all 
algebraic numbers. 


94 SCHOOL ALGEBRA. 


46. The distributive law also holds; for, if 
a(b + c)=ab sen 
then —a(b+c)=—ab—ae, 
and (b +c) (— a) =b(—a) +e(—a). 
Therefore, for all values of a, 6, and e, 
a(b+c)=ab+ae. 


From the nature of division the distributive law which 
applies to multiplication applies also to division. 


47. We have now considered the fundamental laws of 
Algebra, and for convenience of reference we formulate 
them below: 

a+t(b+c)=a+b+e 
a+(b—c)=a+b—e % : (1) 
a—(b+c)=a—b—e 
a—(b—c)=a—b+e | 


(+4) X (+ 6) =+ ab 
Ca) x (— 0) =) 
(—a) x (+ 6) =— ab | 

(— a) x (— 6) =+ ab 


The commutative law: 

Addition atb=b+a . (3) 
Multiplication ab = ba 

The associative law: | 
Addition ee = ae (4) 
Multiplication a(be) = (ab)ec=abe 


INTRODUCTION. Db) 


The distributive law: 
Multiplication a(b+c)=ab-+ ae 


rnees Tae Ae 
_ Division ecicte ita (5) 
a et 
The index law: 
Multipheation iat lia NOeeW nh a Semen mame a can 6) 


These laws are true for all values of the letters, but in (6) 
m and n are for the present restricted to positwe integral 
values. 


48. Value of an Algebraic Expression. Every algebraic ex- 
pression stands for a number; and this number, obtained 
by putting for the several letters involved the numbers for 
which they stand, and performing the operations indicated 
by the signs, is called the value of the expression. 

In finding the values of algebraic expressions, the begin- 
ner must be careful to observe what operations are actually 
indicated. Thus, 


4ameansat+ta+a-+a; that is, 4 xa. 
a*meansaxXaxXaxXa. 
Vabe means the square root of the product of a, }, and. 


/abe means the product of the square root of a by be. 


Nore. The radical sign V before a product, without a vinculum 
or a parenthesis, affects only the symbol immediately following it. 


Va-+ 6 means that d is to be added to the square root of a. 


Va-+6 means that b is to be added to a and the square 
root of the sum taken. 


49, In finding the value of a compound expression the 
operations indicated for each term must be performed before 


the operation indicated by the sign prefixed to the term, 


26 SCHOOL ALGEBRA. 


Indicated divisions should be written in the fractional form, 
and the sign X omitted between a figure and a letter, or 
between two letters, in accordance with algebraic usage. 


Thus, (6 — ec) 2 X e+ 26 should be written 





Notr. The line between the numerator and denominator of the 
fractions serves for a vinculum, and renders the parenthesis un 
necessary. 


If 6=4, and c= — 4, the numerical value is 


A eearac ria eee 
=") 4 96 = 9.4 26 =—1+4 26= 25, 
Spa aly a um cag YL in 


Exercise 2. 


Notz. When there is no sign expressed between single symbois 
or between compound expressions, it must be remembered that the 
sign understood is the sign of multiplication. 


If a=1, b= 2, and .¢=.3, find the valwer 


1. Ta— be. 5. 2a—b+e. 9. /4abe. 
2. ac+b. 6. ab+ be— ae. 10. V6abe. 
3. 4ab—c. 7. P+a+e’. 11. &— 8. 
4. 6ab—b—ec. 8. 2ab— 5d5be’. 12. Ve—ai. 
13. a—2(b+c). 16. 66—10bc+12a+2e. 


14. (a+ 6)+2(c—a). 17. 5¢e+(b—a)- (b+a). 
15. V6bce—(b —c). 18. V62. 

If a=1, b=2, c=8, and d= 0, find the value cf 
19. Ta—be+ 6d. 21. 4ab—cd—d. 
20. ac+b—d, 22. 2a—bte, 


INTRODUCTION. yA 


23. ab+ be — ad. 
24. 2ab— d5bc. 


25. W4abcd. 
26. V4abcd. 


21. 
28. 


29. 


30. 


b—c+d. 
a—2(b+e). 

2b (3 —5e)+(a— 2c). 
2 (a+ by’. 


Exercise 3. 


Remove the parentheses (§ 20), and find the algebraic 


sum of 
(6—2)+(3+)). 
=o) (2—8). 
(—844)—(245). 
—6—(2—3-—1). 
2—(5—7+8). 

8 —(7—5+4). 
10 —(5— 6— 7). 
2—(8—38+4). 
(5 — 10) +(8 — 2). 
—7+(8+2-—4). 


ao 
=) 


fewest )) o= 2, and ¢=— 3, 


21. at+b-+e. 
22. a—b+te. 
23. a—b—ce. 


24. 
25. 
26. 


ah OR (HES Gy: 
Wee pe 5). 
Mees Ae 8), 
Ry by 
Beep |) 
Pe (be 10 = 38). 
= Heh eiea:) Fy 
Homepage dy 4. 
Cre ly aee tt 10), 
eal Seth 1}, 


find the value of 
a—(—b)+e. 
a—(—b)—e. 
(Sa)-P (By (=e): 


CHA Pi heghle 


ADDITION AND SUBTRACTION. 
INTEGRAL EXPRESSIONS. 


50. If an algebraic expression contains only wltegral 
forms, that is, contains no J/edfer in the denominator of 
any of its terms, it is called an integral expression. Thus, 
e+ Tcxv’—c—5ex, 4ax—Ltbcy, are integral expressions, 


Wy 2 = a aad —s . ' . 
but As lt seis is a fractional expression. 
v—ab+ 6 


An integral expression may have for some values of the 
letters a fractional value, and a fractional expression an 
integral value. If, for instance, a stands for $ and 6 for 
1, the integral expression 2a — 506 stands for $—$=1; 


and the fractional expression - stands for 43+2=65., 


Integral and fractional expressions, therefore, are so named 
on account of the form of the expressions, and with no refer- 
ence whatever to the numerical value of the expressions 
when definite numbers are put in place of the letters. 


51, A term may consist of a single symbol, as a, or may 
be the product of two or more factors, as 6a, ab, 5a7be. If 
one of the factors is an arithmetical symbol, as the factor 5 
in 5a°be, this factor is usually written first, and is called 
the coefficient of the term; the other factors are called literal 
factors. 

Nore. By way of distinction, a factor expressed by an arithmeti- 


cal figure is called a numerical factor, and a factor expressed by a 
letter is called a literal factor. 


ADDITION AND SUBTRACTION. 29 


52. Like Terms. ‘Terms which have the same combina- 
tion of “itera/ factors are called like or similar terms; terms 
which do not have the same combination of literal factors 
are called unlike or dissimilar terms. Thus, 5a°be, —7Ta’bc, 
abe, are like terms, but 5a%be, 5ab’c, 5abc’, are unlike 
terms. 


53. A simple expression, that is, an expression of one 
term, is called a monomial. A compound expression, that 
is, an expression which contains two or more terms, is — 
called a polynomial. A polynomial which contains two 
terms is called a binomial, and a polynomial which contains 
three terms is called a trinomial. 


54. A polynomial is said to be arranged according to 
the powers of some letter when the exponents of that letter 
either descend or ascend in the order of magnitude. Thus, 
daz — 4bz?— 6bax-+8b is arranged according to the de- 
scending powers of x, and 8b—6axr—462?+ 8az’* is 
arranged according to the ascending powers of wx. 


55. Addition of Integral Expressions, The addition of two 
algebraic expressions can be represented by connecting the 
second expression with the first by the sign +. If there 
are no like terms in the two expressions, the operation is 
algebraically complete when the two expressions are thus 
connected. 

If, for example, it is required to add m+n—~p to 
a+b-+e, the result will ba+6+e+(m+n—~p); or, 
removing the parenthesis (§ 20),a+d+e+m+n—p. 


56. If, however, there are like terms in the expressions to 
be added, the like terms can be collected; that is, every 
set of like terms can be replaced by a single term with a 
coefficient equal to the algebraic sum of the coefficients of 
the like terms. 


30 SCHOOL ALGEBRA. 


If it is required to add 5a?+4a+3 to 2a@—8a—4, 
the result will be 


2a’— 3a—4+(5a?+ 4a+8) 


=2¢—38a—4+5¢+4a+38 § 20 
= 207+ 5¢—38a+4a—4+3 § 26 
=Ta+a—l. 


This process is more conveniently represented by arrang- 
ing the terms in columns, so that like terms shall stand in 
the same column, as follows: 


20 —8a—4 
5a+4a+38 
T+ a—-l 


The coefficient of a in the result will be 5+ 2, or 7; the 
coefficient of a will be —8+4, or 1; and the last term is 
—4+3, or —l. 

Notre. When the coefficient of a term is 1, it is not written, but 


understood ; conversely, when the coefficient of a term is not writ- 
ten, 1 is understood for its coefficient. 


If we are to find the sum of 2a°—8@b+4ae’?+ 6°, 
a+ 4a7b — Tab? — 26°, — 8a*® + ab — 8ab? — 486°, and 
20+ 20¢b+ 6ab?— 30°, we write them in columns, as 


follows” 2a°— 80% +40bt+ B 
a + 407b — Tab? — 26° 
—8a°+ ab—3a’?—48° 
2a°+ 2076 + 6ab? — 36° 
20+ 407 — 86° 
The coefficient of a in the result will be 2+1—38-442, 
or +2; the coefficient of a7b will be —8+4+1-+2, or +4; 


the coefficient of ab? will be 4—7—8-+6, or 0; and the 
coefhcient of 6* will be 1—2—4-— 8, or —8. 


ADDITION AND SUBTRACTION. dl 


Exercise 4, 


Add 
Ta, 2a, —8a, and — 5a. 


— 


Tay, 2xy, —4axy, and — 5xy. 

4a°b, —3a’b, and — 5a’d. 

day, 4ay, Tax, and — 3az. 

a+6 and a—b. 

xv? —x and 2*— 2’. 

5a? +62—2 and 827—Tx+2. 

3827°—2ry+y and 2’? —2ry4+3y’. 

ax’ + be —4, 3a27—2bx+4, and — 4a2? —2bxa+ 85. 

5a+3y+z2, 8x+2y+ 382, and x—d8y—5z. 

11. 8ab—2a2z*+ 3a'%z, 4ab—6a'?x+5a2’, ab+ aa— az’, 
and az’— 8ab—5a’x. 

12. at —2a° + 8a?—a+7 g 2 at mens +2a¢—a+6, and 
—a —2¢04+2¢— 

13. 38¢0—ab+ac— 30'?+4be—¢, —5a’?—ab—ac+5be, 
—4b6e+ 52+ 2ab, and —40?+ 8? —5bce4+ 2c’. 

amg oe? 4a ty Bat 4 O28 4+ 2? — 5a — 6, 
—4a'+32*-—382°+9x—2, and 22t—a2’?4+2?—2+1. 

15. 3a7°—42°y+2', 5a°?—llay—1222, —Ty+2°y—22’, 
and — 4227+ 7?— 2° 

16. a —2a°+3a7, dt+a’+a, 4at+5a', 2V7+3a—2, 


and — a’? —2a— 83. 


ay 
ie 


17. #& +227 —axy—y’, 22° —382y — 42y — Ty’, 
and 2° — 8 ay"? — 7. 


ne SCHOOL ALGEBRA. 


57. Subtraction of Integral Expressions. The subtraction 
of one expression from another, if none of the terms are 
alike, can be represented only by connecting the subtra- 
hend with the minuend by means of the sign —. 

If, for example, it is required to subtract a+6-+e¢ from 
m+n — p, the result will be represented by 


m+n—p—(a+b+e); 
or, removing the parenthesis, § 20, 
mtn—p—a—b—e. 


If, however, some of the terms in the two expressions are 
alike, we can replace two like terms by a single term. 
Thus, suppose it is required to subtract a’—2a’+2a—1 
from 3a°—2a?+a—2; the result may be expressed as 
follows : 
380° —2¢+a—2—(a— 20+ 2a—1); 
or, removing the parenthesis (§ 20), 
8e0—2V+a—-2—a+2¢0—2a+1 
=38ae—a—2¢4+2¢0+a—2a—-241 
=2a—a—l. 


This process is more easily performed by writing the 
subtrahend below the minuend, mentally changing the sign 
of each term in the subtrahend, and adding the two expres- 
sions. Thus, the above example may be written 


38a°—2a+ a—2 
@&—2a?+2a—1 
2a — a—l 
The coefficient of a® will be 8—1, or 2; the coefficient 
of a? will be —2+ 2, or 0, and therefore the term a? will 


not appear in the result; the coefficient of a will be 1— 2, 
or --1; the last term will be —2-+-1, or —1. 


ADDITION AND SUBTRACTION. 338 


Again, suppose it 1s required to subtract a°+ 4 a*x’— 3 aa 


—4az* from a*2?+ 2a’2?—4azx‘. Here terms which are 
alike can be written in columns, as before: 


a’a? + 2 a?x*® — 4az* 
a + 4a°2 — 8a’2* — 4az* 
— o°— Baz’ + 5a’2’ 

There is no term a in the minuend, hence the coefficient 
of a in the result is O—1, or —1; the coefficient of a’a? . 
will be 1 —4, or —3; the coefficient of a’z* will be 2+8, 
or +5; the coefficient of ax* will be —4-+4, or 0, and 
therefore the term az* will not appear in the result. 


Exercise 5. 


1. From 8a— 46 — 2c take 2a— 36 — 8c. 

2. From 8a—46+ 8c take 2a— 8b—c—d. 

8. From 7a?—9a—1 take 5a°—62—83. 

4. From 22?— 2axv-+ a’ take 2? — ax — a’. 

5. From 4a— 36 — 8c take 2a— 36+4e. 

6. From 52°+ 72+4 take 32?— 7x4 2. 

7. From 2ax+ 3 by +5 take 38ax — 3 by — 5. 

8. From 4a?— 6a04+ 207 take 8¢’+ab-+ 6. 

9. From 407) + 7ab?+ 9 take 8 — 3a0’. 
10. From 5a’ + 6a7b — 8a’ take 6° + 6a’) — 5a’e. 
11. From a — 2 take 0”. 13. From 0? take a? — 0. 
12. From a? — 0’ take a’. 14. From a’ take a? — 6’, 
15. From 2*+ 3axz* — 262? + 38cx—4d 


take 82*+ az®—40?+ 6er+d. 


34 SCHOOL ALGEBRA. 


If A=3e—200+587, C=Te—8ab+58', 
B=9¢—50ab4+30?, D=11ld—38ab— 4B, 


find the expression for 


146. 4A+C+ B+ D. 19. A+ C—B-D. 
17. A—C— B+ D. 20. A—C+ B+ D. 
18. C—A—B+D. 21. A+C—B+D. 


58, Parentheses. From the laws of parentheses (§ 20), we 
have the following equivalent expressions : 


a+(6+ec)=atb+e, «. a+b+e=a+(b6+0e); 
a+(b—c)=a+b—c, ».a+b—c=a+(6—e); 
a—(b+ce)=a—b—c, «.a—b—c=a-—(b+e); 
a—(b—c)=a—b+e, .«.a—b+c=a—(b—-oc); 


that is, if a parenthesis is preceded by the sign +, the 
parenthesis may be removed without changing any of the 
signs of the terms within the parenthesis ; conversely, any 
number of terms may be enclosed within a parenthesis 
preceded by the sign +, without changing the sign of any 
term. 

If a parenthesis is preceded by the sign —, the paren- 
thesis may be removed, provided the sign of every term 
within the parenthesis 1s changed, namely, + to — and — to 
-+; conversely, any number of terms may be enclosed with- 
in a parenthesis preceded by the sign —, provided the sign 
of every term enclosed is changed. 


09, Iixpressions may occur with more than one paren- 
thesis. In such cases parentheses of different shapes are 
used, and the beginner when he meets with a ( or a[ ora 
} must look carefully for the other part, whatever may in- 
tervene; and all that is included between the two parts of 


ADDITION AND SUBTRACTION. 35 


each parenthesis must be treated as the sign before it directs, 
without regard to other parentheses. It is best to remove 
each parenthesis in succession, beginning with the innermost 
one. Thus, 


(1) a0 -=(e--.0) +e] 
=a—[b—c+d+e] 
=a—btc—d—e. 

(2) Pane sacra ti ey 7 |, 


=a—{b—[e—d+et/} 
=a—{b—c+d—e—f} 
=a—b+c—d+e4+f. 


Exercise 6. : 


Simplify the following by removing the parentheses and 
collecting like terms: 


1. a—b—[a—(b—c)—el]. 

m —|n—(p —m)]}. 

20 —fy + [42—(y + 22)]}. 

8a —{2b—[5ce—(8a+4)]}. 
a—{b+[e—(d—b)+a]— 28}. 

32 —[9—(224 7)4+ 32]. 

2a —[y—(%—2y)]}. 
a—[26+(8¢—26)+a]. 

(a—2 +y)—(b—2—y) + (a+b —2y). 

8a —[—4b+(4a —6)—(2a—5d)]. 

. 4c—[a—(26 —3c)+c]+[a—(26—5e—a)]. 
. ©+(y—2)—[B2—2y) +2]+[2—y—2)] 
. a—[2a+(a—2a)+ 2a]—5a—{6a—[(a+2a)—-]}}. 


Oo pets oy HO! lomo ee 


— et 
Oo te St CS 


36 SCHOOL ALGEBRA. 


14. 2x—(8y+2)—}b—(e—b)+e - [a—(e—B)}}. 
15. a—[b+¢—a—(a+6b)—c]4+(2a—b+e). 


‘Nors. The sign — which is written in the above problem before 
the first term 6 under the vinculum is really the sign of the vinculum, 
—b +c meaning the same as —() + ¢). 





16. 10—x#%—j{—x—[2—(*4#—5—2z)}]}. 

17. 2e—f2et(y—z)—824[24—(y—z—2y)—82]+4y}. 
18. a—[b—{—ce+a—(a—b)—c}]+[2a—(6—a)]. 
19. a— jb —[a—(e—b) +e—a—(a—b —c)—a]+ a}. 
20. 5a—{—8a—[8a—(2a—a—b)—a]+a}. 


Exercise 7. 


v 

In each of the following expressions enclose the last three 
terms in a parenthesis preceded by the sign —, remember- 
ing that the sign of each term enclosed must be changed. 


1. 2a—b—38ce—d+38e-—5f. 

x—a—y—b—z-e. 

at+tb—ce+4a—6-+1. 

ax + by + cz + ba — cy + cz. 
38a+2b6+2c—5d—8e—4f. 
x—y+2—5xry—422+ 8 yz. 

Considering all the factors that precede a, y, and z, respectively as 


the coefficients of these letters, we may collect in parentheses the 
coefficients of x, y, and z in the following expression : 


i 


ax — by + ay —az—cz + be =(a + b)a+(a—b) y—(a + edz. 
In like manner, collect the coefficients of x, y, and z in 
‘the following expressions : 
7. ax+t by +cz+ bu —cy + az. 
8. ax+ 2ay+ 4az— bx + By — 3 bz — Qz. 


ADDITION AND SUBTRACTION. 3h 


ax — 2by —5cz—4bz%+ 3cy — Taz. 


. a2+d8ay + 2by — bz —1lex+ 2cy — cz. 

. 4by —8axc — bez + 2be — Tex — Sey — cx — cy— ez, 
- 6az—5by + 38cz — 2bz — Bay + bz — ax + by. 

- 2—by+ 8az—38cy + 2ax— 2mx — 5 bz. 


. 2+ ay —az—acx + bez -mny —y —z. 


Exercise 8. 


EXAMPLES FoR REVIEW. 


. Add 42°—5a’?— 5az’*+ 6a'x, 60°+382°+4a2?+ 2072, 


19 a2?—112?—15a’2, and 102°+-7a@2+5a*®—18 az’. 


. Add 8ab+38a+66, —ab+2a+4b, Tab —4a—8), 


and 6a+126—2ad. 


Note. Similar compound expressions are added in precisely the 


same way as simple expressions, by finding the sum of their coefh- 
cients. Thus, 3(¢—y) + 5(a—y)—2(a—y) =6(#—y). 


3. 
4. 


Add 4(5—2), 6(5 — x), 8(5— x), and —2(5 — 2). 

Add (a+ 6)2+(6b+e)¥Y+(ate)2, (6+¢)2 
+(a+te)y+(a+b)2, and (a+c)2+(a+6)y7 
+ (6+ ¢)2*. 

Add (a+ 6)a+(b+c)y+(e+a)z, (6+e)ze+(e+a)x 
—(a+b)y, and (a+c)y+(a+6)z2—(b6+¢e)z. 

From a’?— 2 take a + 2ax-+ 2”. 

From 8a?7+2azr+ 2’ take a — ax — 2’, 

From 827?—3azxr+5 take 527 + 2ax+5. 

From a? + 3 6’c + ab? — abe take ab? — abc + 6’. 

From (a+ 6)2+(a+ce)y take (a—b)x—(a—e)y. 


. Simplify 7a—{3a —[4a—(5a—2a)]}. 


388 SCHOOL ALGEBRA. 


12. Simplify 8a—fa+6—[a+b+c—(a+b+c+d)}]}. 

13. Bracket the coefficients, and arrange according to the 
descending powers of x 
x? —ax—c'x? — ba + ba? — cx’? + aa? — 2? — cx. 

14. Simplify a’—(6’—¢’)—[&—(e—a’)|+[e—(0 —a’)]. 

15. If a=], b=38, c=5, and d=7, findothesvameso: 
a — 2b—$3e—d—[8a—(5b—c—8d)|—28}. 

16. From 2d+1lla+106—5e take 2e+5a—806. Find 
the value of each of these expressions when a, 8, ¢, 
and d have the values 1, 3, 5, 7, respectively, and 
show that the difference of these values is equal to 
the value of their difference. 

17. If a=1, b=— 3, ¢-=— 5, d=0, find the’ vaineus 
?+26?+ 38+ 4d". 

If a=3, 6=4, c=9, and 2s=a+6 +e findethe 
value of 

18. s(s—a)(s— b)(s—e).’ 

19. s?+(s—a)’+(s—6)?+(s—e). 

20. s’—(s—a)(s —b) —(s — 8) (s—e) —(s—e)(s— a). 

21. Ift=a+2b—3c, y=b+2c— 8a, and z=c+2a 
— 386, show that 7+y+2=0. 

22. Ift=a—2b+3c, y=b—2c+3a, and z=e—22a 
+36, show that a+y+z2=2a+26+4 2e. 

23. What must be added to 2’?+ 5y?+ 32° in order that 
the sum may be 27? — 2”? 

24. What must be added to 5a3— 7a7b+ 8ab? in order 
that the sum may be a® — 2a7b — 2ab? + 6? 

25. If H=50°+3a7)—205, F=3a' —7a’b — 6B, 

G=2¢b—a—B, H=avb—2a'— 38°, 
find the expression for #'—[/’'—(G— A)]. 


CHAPTER III. 


MULTIPLICATION. 


INTEGRAL EXPRESSIONS. 


60. The :aws which govern the operation of multiplica- 


tion are formulated as follows: 


ab = ba 
ax (oc) = (ab) X c= abe 


a(b+c)=ab-+ac | 


a(b—c)=ab—ac 
axa =a" . 
aX (+4) =-+ab 
aX (— 6) =— ab 
(—a)xb =—ab 
Ge) x (-— 6) = +05 


61. Multiplication of Monomials. 


§ 47 


The commutative law. 


The associative law. 
The distributive law. 


The index law. 


The law of signs. 


When the factors are 


single letters, the product is represented by simply writing 
the letters without any sign between them. Thus, the prod- 


uct of a, 6, and ¢ is expressed by adc. 


62. The product of 4a, 56, and 3c is 
4ax5bx3c=4X5 xX 38abe = 60abe. 


Norse. We cannot write 453 for 4x 5x3 because another mean- 
ing has been assigned in Arithmetic to 453, namely, 400 + 50 + 3. 
Hence, between arithmetical factors the sign must be written. 


40 SCHOOL ALGEBRA. 


63. The product of ab and a’? is 


Cb xX &b == eabh = FOV aa 


64, To multiply one monomial by another, therefore, 


Find the product of the coefficients, and to this product 
annex the letters, ging to each letter in the product an index 
equal to the sum of its indices in the factors. 


Notre. The beginner should determine first the sign of the product 
by the law of signs, and write it down; secondly, after the sign he 
should write the product of the coefficients; and lastly, each letter 
with an index equal to the sum of its indices in the factors, 


65. We may have an index affecting an expression as 
well as an index of a single letter. Thus, (adc)? means 
abe X abe, which equals aabdbce, or a’b’c?,_ In like manner, 
(abe)*=a"b*c"®. That is, 


The nth power of the product of several factors is equal to 
the product of the nth powers of the factors. 


66. By the law of signs, we have 
(= a). X. 2) = Had, 


and (+ ab) x (—e) = — abe, 
that is,  (—a) x (— 4) x (—¢e) =—abe; 
and (— abc) x (— d) =+ abcd, 


that is, (— a) X (— b) X (—e) X (—d) = + abed. 
It is obvious, therefore, that 


The product of an even number of negative factors will 
be positive, and the product of an odd number of negative 
factors will be negative. 


67. Polynomial 


> 
MULTIPLICATION. 41 


s by Monomials. We have (§ 47), 


a(b + ¢)=ab + ac. 
In like manner, 


a(b—c+d—e)=ab—ac+ad— ae. 


To multiply a polynomial by a monomial, therefore, 


Multiply each 


term of the polynomial by the monomial, 


and add the partial products. 


oO fF WO DN & 


Exercise 9. 


Find the product of 


Te and 58. 6. Tab and 8ac. 
3x and 8y. 7. —2aand 7a*a’y. 
8a’ and 6a’. 8. —38a’b and — 8ab’. 
3a and 24°. 9. —5mnp’ and —4m’n*p’. 
2mn and 3m’n. 10. — 8a’, —20’, and —3 ab. 
11. —2277y, xy’, and —3x’y. 
12. —3a*y, —2a7b, and — a’y'a°b". 
13. 5a+36 and 2a’. 
14. ab—be and 5avbe. 
15. ab —ac— be and abc. 
16. 6a°b — Ta’b’c and a’b’c. 
17. 2+ b—e' and a®bc’. 
18. 5a’? — 367+ 2¢ and 4ab'e’. 
19. abe— 38a°b2 and — 2ab’e. 
20. —xy2?+ a2’y*z and — xyz. 
21. —2m’np* —mnp’ and — m’np. 


os 
42, SCHOOL ALGEBRA. 


299/799 


22. x—y—szand — 32°y'2". 
23. —32’ and w+ 27’ —z. 
24. 3x—2y—4 and 52’. 


68, Polynomials by Polynomials. If we have m+n-+p 
to be multiplied by a+ 6-+¢, we may substitute I for the 
multiplicand m+n-+ p (§12). Then 


(a+b+c)M=aM+bM+ cM. § 28 


If now we substitute for JZ its value m-+7-+- p, we shall 
have 


a(m+n+p)+b(m+n+p)+e(m+n+p) 
=am-+an+ap+bm-+ bn+ bp+em-+en+ep. 
That is, to find the product of two polynomials, 


Multiply every term of the multiplieand by each term of 
the multiplier, and add the partial products. 


69. In multiplying polynomials, it is a convenient ar- 
rangement to write the multiplier under the multiplicand, 
and place like terms of the partial products in columns. 


ak 5 aes 7606 
Dee teb 
15a@— 18ab 

— 20ab + 246? 


15a? — 38ab + 240° 


We multiply 5a, the first term of the multiplicand, by 
3a, the first term of the multiplier, and obtain 15a’; then 
— 66, the second term of the multiplicand, by 3a, and ob- 
tain —18ab. The first line of partial products is 15a’ 
—18ab. In multiplying by —40, we obtain for a second 
line of partial products —20ab+ 240’, which is put one 
place to the right, so that the lke terms —18ad and 


MULTIPLICATION. 43 


—20ab may stand in the same column. We then add the 
coefficients of the like terms, and obtain the complete prod- 
uct in its simplest form. 


(2) Multiply 424+ 3-4 52°— 62’ by 4—62?—5xz. 


Arrange both multiplicand and multipher according to 
the ascending powers of z. 


8+ 4a2+ 52°7— 62° 
4— 5br— 6x 


12+ 162+ 202? — 242° 
— 1ldx2 — 202? — 252° + 302* 
— 1827 — 242° — 302* + 362° 


12+ #—182?— 732’ + 362° 


(3) Multiply 1+ 22+ z*— 327 by a —2— 22. 
Arrange according to the descending powers of z. 


z*—327+2r+1 


| 
v—32°4+2et+ 2 
—22° + 62° —42°—227 
— 2x* Gar 4a S22 
z'’— 52° + 7a’?+227—62—2 


(4) Multiply a?+ 8’+ c?— ab —be—ac by a+b+e. 
Arrange according to descending powers of a. 


@—ab—act+ B-~ be+ ¢ 


a+ 6+ ¢ 
&—ab—aetabl’?— abetace 
+ a7b —ab*’— abe + 63 — Be + be? 
+ are — abe—ac’ + b6%e—b’?+é 


a — 8abe + 6° + 


44 SCHOOL ALGEBRA. 


Norr. The student should observe that, with a view to bringing 
like terms of the partial products in columns, the terms of the multi- 
plicand and multiplier are arranged in the same order. 


70. A term that is the product of three letters is said to 
be of three dimensions, or of the third degree. In general, 
a term that is the product of 7 letters is said to be of x 
dimensions, or of the nth degree. Thus, 5abe is of three 
dimensions, or of the third degree; 2.a7’c’, that is, 2aabbce, 
is of six dimensions, or of the sixth degree. 


71. The degree of a compound algebraic expression is the 
degree of that term of the expression which 1s of haghest 
dimensions. 


72; When all the terms of a compound expression are of 
the same degree, the expression is said to be homogeneous, 
Thus, 2°+ 32°y + 3ay’+ 7° is a homogeneous expression, 
every term being of the third degree. 


73. The product of two homogeneous expressions 1s homo- 
geneous. For the different terms of the product are found 
by multiplying every term of the multiplicand by each term 
of the multiplier; and the number of dimensions of each 
partial product is the sum of the number of dimensions of 
a term of the multiplicand and of a term of the multiplier 
counted together. Thus, in multiplying v+0?+¢— ab 
—be—ace by a+6+ ce, Example (4), each term of the mul- 
tiplicand is of two dimensions, and each term of the multi- 
plier is of one dimension; we therefore have each term of 
the product of 2+-1, that is, three dimensions. 

This fact affords an important test of the accuracy of the 
work of multiplication with respect to the hteral factors ; 
for, if any term in the product is of a degree different from 
the degree of the other terms, there is an error in the work 
of finding that term. ; 


MULTIPLICATION. 45 


74. Any expression that is not homogeneous can be 
made so by introducing a letter, the value of which igs 
unity. ‘Thus, in Example (3), the expressions can be writ- 
ten z*—3 a’x’ + 2a%x-+- at and 2° —2a’x—2a*. The prod- 
uct will then be 2’ — 5a7a°+ Tata + 2a°x? — 6a®x — 2a’, 
which reduces to the product given in the example, by 
putting 1 for a. 


75. It often happens in algebraic investigations that 
there is one letter in an expression of more importance 
than the rest, and this is therefore called the leading letter. 
In such cases the degree of the expression is generally called 
by the degree of the leading letter. Thus, a?x?+bx-+¢ 1s of 
the second degree in x. 


Exercise 10. 


Find the product of 


1. 2+10 and 2+ 6. 12. 2x—-3 and z+ 8. 

2. x—2and x — 3. 13. x—T7 and 2x—1. 

3. x—38 and 2+ 5. 14. m—nand 2m+1. 

4. +3 and x— 3. 15. m—aandm-ta. 

5. a—Illand2a—l. 16.:382+7 and 2x —8. 

6. —x+2and—2—3, 17. 54a—2y and Ba + Qy. 
7. —x—2 and x — 2. 18. 82—4y and 227+ 3y. 
8. —x+4andx—4. 19. 2?+ 7’ and 2° — 7’ 

9. —x+7 and 2+ 7. | 20. 227+ 3y’ and wv’ + 7’. 
10. <—Tandx+7. 21. at+y+zand x—y+z. 
11. -~—8 and 22+ 3. 22. «+2y—z and —y+2z, 


23. 2 —ay+y’ and 2+ 2y+y7/’. 
24. m’—mn+n? and m+n. 


25. M+tmn+n and m—n. 


46 


st Oo Tm FPF DO WO & 


SCHOOL ALGEBRA. 


26. a@— 3ab+6' and a?— 3ab — 8. 

27. a —Ta+2 and a&’—2a+3. 

28. 2v°—d3ay+4y’ and 82°+ 42y — dy’. 

29. v+ay+y and x2 —xrz2—2’. 

30. Y+y7Y+2—ay—az—yzande«+y+z. 
31. 4a®—10ab + 250°? and 56+ 2a. 

32. w+t4y andy’ +4z. 

33. 2+ 22y4+8 and y’ + 2ay — 8. 

34. V?@+6+1—ab—a—banda+1+0. 
35. 38a7— 27+ 52 and 827+ 27 — 32’, 

36. v2 +y' + 2xy—2x—2y—1 anda+y-—l. 
37. a®+ 2a"! —3a™’— 1 anda-+1. 

38. a"—4a"'+ 5a°? +a" anda—l. 

39. ait! — 4a**4+ 2a"! — a" and 2a°— a’ +a. 


40. 2*—y"" and a®*+y""". 


Exercise 11. 

Simplify : 

(a+6+c)(a+6—c)—(2ab—c’). 

(m+n) m—[(m—n)’—n(n—m)]. 

[ae — (a — 6) (6+ e)] — b[b —(a—c)]. 

(% —1)(a — 2) — 3x2 (x#+8)+2[(e+4+ 2)(@+ 1) — 3]. 
4(a— 3b)(a + 36)—2(a— 66) —2 (a? 4 68"). 
(etyt 22—2yt2—x)—y(@te—y)—2@+y—2). 


5 [(a—b) x—cy]—2[a(w— y)— be] 
—[S8ax—(S5e—2a)y]. 


CHO ABA Pw VY 


10. 
Ai. 
12. 
13. 
14. 
15. 
16. 
17. 
18. 
19. 
20. 


a 
22. 
23. 


MULTIPLICATION. 47 


Exercise 12. 


EXAMPLES FoR REVIEW. 


Multiply 
z’*—x2x—19 by #4 22—83. 
1+22+2? by 1—a2'+ 22°— 382. 
22°+2+4+32 by 24-3274 22°. 
382°+5—42 by 8+ 62?—7z. 
ate—y by x#—y'+ zy. 
38+ 72°— 5x by 82°—62—102°+ 4. 
6? + 6ab?—4a°b by 2a7b — ab? — 8a'. 
x’+ax—b by #+52—4. 
e—me’+ne+r by 2+cx4d. 


x’—(a+b)x+ab by x—c. 

a + xy + xy? + y* by y— x. | 

427+ 9y* — bay by 42°+ 97’ + bay. 

x —32?+5 by 2°+4. 

a — xy? +y* by at vy? + y'*. 

2a°—32°+42?—5 by 2’ —8. 

am —amy™ +a" by a™-+y"™. 

a*—a*+a*—1 by a*¥+1. 

a + B+ P+ ab — ac — be by a+b+e. 

Simplify (a — 26) (6—2a)—(a—386) (46 —a)+ 2ad. 

If a=0, 6=1, and c= — 1, find the value of 
(a—b)(a—c)+e¢(8a—b—c)+ 2ace—(a—e)4+20. 

[22+ y+ (@—2y)'][Be—2y)P— (2x —8y)]. 

a’ (b—c)—b’? (a—ec)+¢ (a—b)—(a—b) (a—c) (6—c). 

(2a—6)?+26(a+b)—38a’?—(a—b)’+ (a+) (a—B). 


CHAPTER IV. 


DIVISION. 
INTEGRAL EXPRESSIONS. 


76. The laws for division are expressed in symbols, as 
follows : 

+ab+(+a)=+6 

—ab+(+a)=—6b 


L f signs. 44 
abst ( Sey aw of signs § 
—ab+(—a)=+6 
b+e_8,¢ | _ Distributive law. § 47, (5) 

a a a 


77. The dividend contains all the factors of the divisor 
and of the quotient, and therefore the quotient contains the 
factors of the dividend that are not found in the divisor. 


Thus, woe Si a rae a Ste 
Cc a ——- a 








78. If we have to divide a® by a’, a® by a‘, a* by a, we 
write them as follows: 








a  aaaaa pe 
>= =aaa=a=a>’, 
a au 
a’ aaaaaa <a oe 
BV es —-qdaa =—-a=—-a ; 
a“ aaaa 

| 


a  aaaa 


a a 





DIVISION. + 49 


79. To divide one monomial by another, therefore, we have 
the following rule : 


Duwide the coefficient of the dividend by the coefficient of 
the dwisor (observing the law of signs), and subtract the 
index of any letter in the divisor from the w dex of that letter 
in the dividend. 





—_—_—_—__. =e 


Thus, Bese Sy so Tab: SOU Oe 
a 





8 q'"-! is l : Leakey? 3ee 
‘abso p—6, r—1 
38a REL ID AL, 


2 n : n 
a Nore. Since“ =1, and also by the rule above given, % = a-™ 
a” ? y 8 ; qn 


=a, it follows that a®=1. Hence, any letter which by the rule 
would appear in the quotient with zero for an index, may be omitted 
without affecting the quotient. 


80. To divide a polynomial by a monomial, we have, by the 
distributive law, the following rule : 


Divide each term of the dividend by the divisor, and add 
the partial quotients. 


8ab-+ 4ac—6ad_8ah_, 4ac_ 6ad 





Thus, 


2a Zi 2a 2a 
= 461-2¢— 3d. ; 
9a‘h?s—120°b2? —3.a7x pee a‘b’s l2a*bs?  3a%e 
3a°x 83 a’7x 8 ax Sar 


= Sob? — 4ab2— I, 


Gait a 47°" 6 gintl 4 an 


= —- mae Sb St ae 
9 gin 9 gin-l y gil 








Nott. Here we have 4n+1—(2n—1)=4n+1—-2n41=2n+2, 
and 3n—(2n—1)=3n—2n+1=n +1, as indices of & in the first 
and last terms of the quotient respectively. 


50 


j—_ 


et aa oe 
Det 


CM NS TP w W 


SCHOOL ALGEBRA. 


Exercise 13. 


Divide 
3a° by a’. 
—422" by 62°. 
— 352 by 52. 
— 62 by — 8z. 
202'y? by — 4xy’. 
— 212+ by — 72%. 
28 a'b® by — Tab’. 
— 252°b? by — 526. 
— 24m'‘n® by —4m?n’. 


— 35 p97? by 5pq’. 


- —167°s° by — 47°. 


. 28m™n" by 4min™. 


13. 
14. 
15. 
- 188'x" by —9 5%". 

- 256.2°4%2" by 82%y7z. 

. —50a°b* by — 10a°b. 
. 84 a'y® by 14277. 

. —80a'x’2 by — 622. 


23. 
24. 


abx’ by abz. 
— 2a°ba* by ax. 
4abx’ by —azx’. 


- x -+ 2ay by a. 
. &—2ab by a. 


4a® — 8x’ by 22°. 
—62°— 2x by —2z. 


25. — 8a’ — 16a” by — 8a’. 

26. 27a* — 36a? by 9a’. 

27. —380a'+ 20a’ by — 10a°. 
28. —122°y*— 42° by — 4277’. 


29. —32x"2"— 622° by —32°2*, 
30. 3a°b*c' —9arbdic’ by 3.a°b%c’. 
31. 2?— xy — xz by —z. 

32. 3a>--6a7b — 9ab® by — 8a. 
38. ay? — 2*y* — ay" by 277’. 
34. a’b?c — a’b*ce — a’bc’ by abe. 
35. 8a°—4a*b — 6al” by — 2a. 


36. 5m'n — 10mn? — 15 mn by 5mn. 


DIVISION. 51 


81. To divide one polynomial by another. 


If the divisor (one factor) = a+6b-+e, 
and the quotient (other factor)= ntpt+gq, 
an+bn+en 
then the dividend (product) =} -+ap+bp+ep 
‘tag+bq-+ cq. 


The first term of the dividend is an; that is, the product ° 
of a, the first term of the divisor, by , the first term of the 
quotient. The first term ” of the quotient is therefore 
found by dividing an, the first term of the dividend, by a, 
the first term of the divisor. 

‘If the partial product formed by multiplying the entire 
divisor by » be subtracted from the dividend, the first term 
of the remainder ap is the product of a, the first term of 
the divisor, by p, the second term of the quotient; that is, 
the second term of the quotient is obtained by dividing the 
first term of the remainder by the first term of the divisor. 
In like manner, the third term of the quotient is obtained 
by dividing the first term of the new remainder by the first 
term of the divisor; and so on. Hence we have the fol- 
lowing rule: 


Arrange both the dindend and divisor in ascending or 
descending powers of some common letter. 

Divide the first term of the dindend by the first term of 
the dwisor. 

Write the result as the first term of the quotient. 

Multiply all the terms of the disor by the first term of 
the quotrent. 

Subtract the product from the dividend. 

If there be a remainder, consider vt as a new dividend 
and proceed as before. 


52, SCHOOL ALGEBRA. 


82. It is of fundamental importance to arrange the divi- 
dend and divisor wm the same order with respect to a com- 
mon letter, and to keep this order throughout the operation. 

The beginner should study carefully the processes in the 
following examples : 


(1) Divide 2?+ 182+ 77 by «+ 7. 


e+ 18a"+ Ae 

e+ Tx zx+1l1 
lla+77 
llz+77 


Nore. The student will notice that by this process we have in 
effect separated the dividend into two parts, 2? + 7a and lla +77, 
and divided each part by «+7, and that the complete quotient is 
the sum of the partial quotients x and 11. Thus, 

4+ 1844+ 77 = 02+ 724+ 1le+ 77 =(0? + 7x) + (112+ 77); 
elisa tT eae Lee ee 


il ee ea | 
“+7 e+7 e+7 


(2) Divide a—2ab+0 by a—O. 
a—2ab+B'?la—b 
a’— ab a—b 
— ab+ 
— ab+P 


(3) Divide 4a*x?— 4a7a* + 2§—a® by 2?— a’. 


Arrange according to descending powers of z. 


ti —4ect+ 4a'r*?—a’le?— oe? 
x*— a'r | xt — 8a’z*+a* 
— 8a7x'* + 4a‘z?— af 
— 8a'x*+ 38 a‘2? 
a‘? — af 
a‘z*? — a® 


DIVISION. 53 


(4) Divide 22073? + 150* + 3a* — 100°) — 22.a8® 
by a&’ +30? — 2ab. 
Arrange according to descending powers of a. 


8a*t — 10.a°d + 22070? — 22.00? +1584] a’'—206+386 
8a'— 6a%d+ Ya’? peace 
— 4a°b + 1807)? — 22ab 
— 40°6+4+ 8a’b? — 12ab’ 
5a’b? — 10 ab? + 15 0 
5b? — 10 ab? + 150+ 


(5) Divide 52°79—xv+1—382* by 1+ 82? —2z. 
Arrange according to ascending powers of 2. 
l— 24+52'—82'*|1—224+327 
1—224+ 382 1+ z#«- x 
x — 327+ 52° — 82* 
xa— 224+ 382° 
— #+22° —32* 
— #422? — 32" 


(6) Divide a+ 7° + 2— d3ayz by r+ y+4+2z. 
Arrange according to descending powers of x. 


v—d3yuatyteletytez 
Bt yn? + 2H eB are 
— ye — 20° — 8y2e + ye +2 
ye Yc yee 
—204+. wa —Qyze+y+2 
— 22" — Yeu — 22 
yx— yaet+ertyte 
pac ++ yz 
— yee+ern—yet2 


eee — Yt — Ye" 
2a + 2” + 2° 


oe ye +2 


54 SCHOOL ALGEBRA. 


(7) Divide 4a7*? — 8007+ 19a*!+ 5a*”’+ 9a** 
by a* ?— Ta®*+ 2a — 8a**, 


4a7t1_30a7+19a7"4+ 5a® 49a" oa a 
4a7_98 a7 8a*—12a*” 4a'—2a°— 8a 
— 2a*+lla®* "+17 a*?+9 a*“* 
— 2a*+14a*""— 4a**4+6a** 
—' 30°19] o*_ Ga" sas 
—3 a? 1-21 a? 7-6 a* 9 a" 





Note. We find the index of a in the first term of the quotient by 
subtracting the index of a in the first term of the divisor from the 
index of a in the first term of the dividend. Now (# + 1)—(«#—3) 
=e2+1—x2+3=4. Hence 4 is the index of a in the first term of 
the quotient, In the same way the other indices are found. 


Exercise 14. 

Divide 
1. @+7a+12bya+4. 6. 42?+122+4+9 by 22+3. 
2. a@—5a+6bya—38. 7%. 62°—11x+4+4 by 8x4—4. 
3. @+2xry+y by x+y. 8. 8x2°—-l0axr—8a? by 42+. 
4. «—2xey+ybyx—y. 9. 8a°—4a—4 by 2—a. 
5. a —y by x—y. 10. a°— 8a—3 by 3—a. 

11. a*+11la’?— 12a—5a®+6 by 8+ a?— 8a. 

12. ¥°—9y+7>—l6y—4 by 7+4+4y. 

13. 36+ m*—138m by 6+ m+ 5m. 

14. 1—s—3s’—s° by 14+ 2s+4+ s*. 

15. b°— 26+ 1 by &—26+1. 

16. 24 227y°+ 9% by v—2xry+ 3y’. 

17. a2 + 6° by at—a’d 4 ab? — ab} + bt. 

18. 1+52°—62* by 1—2x+ 382". 


19. 
20. 
21. 
22. 
23. 
24. 
25. 
26. 
27. 
28. 
29. 
30. 
31. 
32. 
33. 
34. 
35. 
36. 
37. 
38. 
39. 
40. 
41. 
42. 
43. 


DIVISION. oO 


82’°y? + 9y*+ 1l6x* by 427+ 3y’— 4zy. 
e+yte+ 3a’y+3uy by e+y+z. 
J+ 6+ c¢— sabe by a+b+e. 
+ 8y+2—6ayz by +474 2—4x2—2ary—2yz. 
227—3y'+ xy —xz—4yz2—2 by 2a+3y+2. 
v—y—2yz—-2 by x+y+z. 
a+ ay + y! by 2+ ay + y. 
xt — 9a? + 122—4 by 2? +32—-2. 
y — 2y'— 6 +4y? + 18y+6 by ¥+3y4+3y4+1. 
y—dyz' + 42° by y— y2— 22’. 
—Ay?— 924 1l2yz by x+2y — 82. 
2 —41%—120 by #4 42+ 5. 
a—3+52+a—42' by 8-—2x—2’. 
6 — 2a*+ 102°—1lz’+2 by 4x—3-— 22’. 
1—62°+ 52° by 1-244 2”. 
a+ 81+92? by 8x—2°—9. 
e—y by 2#+2y+y’. 
a+y®> by 2+ 7’. 
v+ax+at by 2?—axr+a’. 
v7 —20?+ab—8e+ 7Tbe+2ac by 8c+a—b. 
ab + 2a?—30b’—4be—ac—e by c+ 2a4 36. 
15a*+ 10a*x+ 4072? + 6az*?— 32* by 3a?+ Zax — 2’. 
a®— 86>—1—6ab by a— 26-1. 
on" — 3 xy" +. 8.2%" — y™ by 2*—y". 
amtnhn — 4 gmtn-1h% _ 97 qmtn—2h 3m 4. 40 qntn 3h 
by a”-+ 3a”'b" — 6a™ 6". 


56 SCHOOL ALGEBRA. 


83, Integral expressions may have fractional coefficients, 
since an algebraic expression is integral if it has no detter in 
the denominator. The processes with fractional coefficients 
are precisely the same as with integral coefficients, as will 
be seen by the following examples worked out: 

(1) Add 4a?—4ab+10’, and £a0°+ 2ab — 30’. 

4@—4tab+1i0 
gv+ 2ab— 3h 
4a°+4ab—10 
(2) From 4a’?—+4ab+40? take ta’—4ab+ 30’. 
ta’—tab+ 10 
4a0°—tab+ 20° 
tv? +4ab— 750 
(3) Multiply 4a?—4ab+40° by 4a — 20. 

ta@—tab + 10 

da —26 

ta’—ta’b+ tab’ 

—4tvb+ 2ab?—16' 


ta —1a’b + 28 ab?— 18° 


(4) Divide 20°+ 4. d?— 448d — 5, d* by 8b $d 


25% — 448 3°d + 115d? — 5d’ 3b — 3d 

2h3— 20b%d eo 
— ,0d+1Lbd?— 3d’ 
— 2 bd+ bd? 


8 hd? — 5, d? 
3b? — 5d? 


oA PT RP w 


DIVISION. 


Exercise 15. 


Add 407 + 40?ct+ and — 3,0°b — 10% 4. 
From 32’+ 3ax— 4a’ take 22°— 8ax2—1a’. 
From +y—3a— $a+40 take ty +1a—2z. 
Multiply $¢?>—4ce—4 by 4°—4e+Hh. 


Multiply 42—42?+412° by da+42°+ 12%. 


57 


Multiply 0.5m‘ —0.4m'n + 1.2m?n? + 0.8mn'—14nt - 


by 0.4m? —0.6mn — 0.8 n’. 
Divide 5% at—{a*b + 42070’? +1ab*® by 3a4+40. 
Divide —4d°+ d’— $1d*+%d* by —3da’+ 2d. 


Exercise 16. 


EXAMPLES FOR REVIEW. 


. Find the value of 2° + ¥°+ 2— 382yz, if x=1, y= 


and 2=— 3. 


2, 


Find the value of V2c —a, and of V2c—a, if b=8, 


¢=9, and a= 23. 
Add a’b — ab’+ 6° and a’ — 40° + ab’ — 30°. 
Multiply a”— a™b™+ 6 by a”™+ b”. 
Multiply 40°** + 6a™t*+ 9a? by 2a"™*— 3a’. 
Divide 2° + 87? — 1252+ 30xyz by 7+ 2y — 5z. 
Simplify -(# — a)? — (a — 6)? —(a— 6) (a+ 6 — 82). 


Find the coefficient of w in the expression 


zta—2[2a— b(e—2z)]. 


Multiply A gimtin-1 _ Tam in? 5 gn tim—2 by 5 gman 


58 


10. 
sin 
12. 
13. 
14. 
15. 
16. 


17. 
18. 


19. 


20. 


21. 


22. 


23. 
24. 
25. 
26. 
21 »- 


SCHOOL ALGEBRA. 


Divide &d*™— cd *— ci"d* * by eo eee 

Divide my" — m'tyt™ + m?-*y"—* by m*y" 4, 

Divide ait” —a'+ a? by a? *t¥. 

Divide a ha by eee 

Divide:4? — 4° + oy? by at 

Divide 22"— 62°"y" + 6a"y™ —2y™ by u™—y”. 

Divide 2° — 2aa?+ ava — abe —b’x# + a@7b+ ab 
by 2?—ax-+ bx — ab. 

Divide 2+ 2"+1 by a™—a"+1. 


Divide 3 a™t?— 4g™6_— 12 q+ _ 9 gmt 
by gimt4 an. 38 gms. 


Divide 62> — 182° + 182" — 1327 =p 
Dyt2o on ee 


Divide 12a"? — a”? — 20a"! + 19a"— 10a"! 
by 4a"— 3a" + 2a’. 7 


Arrange according to descending powers of the fol- 
lowing expression, and enclose the coefficient of each 
power in a parenthesis with a minus sign before each 
parenthesis except the first : 


x — 2 be — a? — ax — av’? — cx — ae — bcx. 


Divide 1.2a‘v — 5.494.452? + 4.8 072° + 0.9 aa*— 2° 
by 0.6 a% — 22”. 


Multiply 4a? —4ab+10° by 4a +210. 

Multiply ¢a?+. ab +30 by 4a—40. 

Divide 1a? +7A,a0?+ 70° by 4a+ 40. 

Subtract £2’ + 42y+147’ from 42?—txy+ 47. 
Subtract 2+ 1ay—ty from 22?—tay+y’. 


CHAPTER V. 
SIMPLE EQUATIONS. 


84. Equations. An equation is a statement in symbols 
that two expressions stand for the same number. Thus, . 
the equation 82-++2=8 states that 32-+ 2 and 8 stand for 
_ the same number. 


85. That part of the equation which precedes the sign 
of equality is called the first member, or left side, and that 
which follows the sign of equality is called the second mem- 
ber, or right side. 


86. The statement of equality between two algebraic 
expressions, if true for all values of the letters involved, 
is called an identical equation; but if true only for certain 
particular values of the letters involved, it is called an 
equation of condition. Thus, (a+ 6)?=a’?+ 2ab+ 0’, which 
is true for all values of a and 34, is an zdentical equation ; 
and 32+2=8, which is true only when z stands for 2, is 
an equation of condition. 

For brevity, an identical equation is called an identity, 
and an equation of condition is called simply an equation. 


87. We often employ an equation to discover an unknown 
number from its relation to known numbers. We usually 
represent the unknown number by one of the /ast letters of 
the alphabet, as x, y, 2; and, by way of distinction, we use 
the first letters, a, }, c, etc., to represent numbers that are 
supposed to be known, though not expressed in the number- 


60 SCHOOL ALGEBRA. 


symbols of Arithmetic. Thus, in the equation ax+b=c, 
x is supposed to represent an unknown number, and a, 3, 
and ¢ are supposed to represent known numbers. 


88. Simple Equations. An integral equation which con- 
tains the first power of the symbol for the unknown number, 
x, and no higher power, is called a simple equation, or an 
equation of the first degree. Thus, av-+6=c is a simple 
equation, or an equation of the first degree a wx. 


89. Solution of an Equation. To solve an equation is to 
find the unknown number; that is, the number which, when 
substituted for its symbol in the given equation, renders the 
equation an identity. This*number is said to satesfy the 
equation, and is called the root of the equation. 


90. Axioms. In solving an equation, we make use of 
the following axioms: 


Ax. 1. If equal numbers be added to equal numbers, 
the sums will be equal. 


Ax. 2. If equal numbers be subtracted from equal num- 
bers, the remainders will be equal. 


Ax. 8. If equal numbers be multiplied by equal numbers, 
the products will be equal. | 


Ax. 4. If equal numbers be divided by equal numbers, 
the quotients will be equal. 


If, therefore, the two sides of an equation be increased by, 
diminished by, multiplhed by, or divided by equal numbers, 
the results will be equal. 

Thus, if 8a = 24, then 84+4= 2444, 8a—4=24—4, 
4x 8x4=4~x 24 and 84#+4= 24 +4. 


91. Transposition of Terms. It becomes necessary in solv- 
ing an equation to bring all the terms that contain the 


SIMPLE EQUATIONS. 61 


symbol for the unknown number to one side of the equation, 
and all the other terms to the other side. This is called 
transposing the terms. We will illustrate by examples: 


(1) Find the number for which x stands when 
162—11=72+ 70. 


The first object to be attained is to get all the terms 
which contain x on the left side of the equation, and all the 
other terms on the right side. This can be done by first’ 
subtracting 72 from both sides (Ax. 2), which gives 


92—11= 70, 
and then adding 11 to these equals (Ax. 1), which gives 
9xz= 81. 


If these equals be divided by 9, the coefficient of x, the 
quotients will be equal (Ax. 4); that is, x= 9. 
(2) Find the number for which z stands when +-6=a. 
The equation is z+b=a. 
Subtract 5 from each side, x+b—b=a—Bb., (Ax. 2) 
Since +6 and —6 in the left side cancel each other 
(§ 36), we have zs=a—b. 
(3) Find the number for which x stands when w~- b=a. 
The equation is t—b=@. 
Add +6 to each side, 2—b+b=a+b. (Ax. 1) 
Since —b and +6 in the left side cancel each other 
(§ 36), we have x=at+b. 
(4) What number does w stand for when ax+b=cr+d? 


This is the general form which every simple equation in 
x will assume when the like terms on each side have been 


62 SCHOOL ALGEBRA. 


collected. In this equation x represents the unknown num- 
ber, and a, 6, c, d represent known numbers. 
If now we subtract (Ax. 2) cx and 6 from each side of 
the equation, we have 
ax —cu=d—b; 


or, bracketing the coefficients of 2, 





(a—c)x=d—b, 

Whence, dividing both sides by a —e, the coefficient of z, 
we get p asina 
Li = 
a—c 


92, The effect of the operation in the preceding equa- 
tions, when Axioms (1) and (2) are used, is to take a term 
from one side and to put it on the other side with its sign 
changed. We can proceed in a like manner in any other 
case. Hence the general rule: 


93. Any term may be transposed from one side of an equa- 
tion to the other provided us sign is changed. 


94, Any term, therefore, which occurs on both sides with 
the same sign may be removed from both without affecting 
the equality. 


95. The sign of every term of an equation may be 
changed, for this is effected by multiplying by — 1, which 
by Ax. 3 does not destroy the equality. 


96. Verification. When the root is substituted for its 
symbol in the given equation, and the equation reduces to 
an identity, the root is said to be verified. We will illustrate 
with examples: 

(1) What number added to twice itself gives 24? 

Let x stand for the number ; 


SIMPLE EQUATIONS. 63 


then 22 will stand for twice the number, 
and the number added to twice itself will be x + 2-2. 
But the number added to twice itself is 24; 


+ 22— 4. 
Combining x and 22, - 3a = 24. 
Divide by 38, the coefficient of z, x= 8 (Ax. 4) 
The required number is 8. 
VERIFICATION. 2+ 2a = 24, 
8+2x 8= 24, 
8+16= 24, 
24 = 24. 


(2) If 4% —5 stands for 19, for what number does x 
stand ? 


We have the equation Az —5=19. 
Transpose — 5 to the right side, 42 =19+ 5. 
Combine, 44 = 24. 
Divide by 4, x= 6. (Ax. 4) 
VERIFICATION. 474—5=19, 
4x6—5=19, 
24—5=19, 
19 = 19. 


(3) If 8¢—7 stands for the same number as 14 — 4z, 
what number does x stand for? 


We have the equation 82—T=14—A4z. 
Transpose 4x to the left side, and 

7 to the right side, 82+427=—14+ 7, 
Combine, Wher ered 


Divide by 7, e= 38, 


64. SCHOOL ALGEBRA. 


VERIFICATION. 82—7=14—4z2, 
8xX3—T=14—4~x 3, 
2 = 2. 
(4) Solve the equation (a — 3) (— 4) =2(# — 1) — 30. 
We have (% — 8) (a — 4) =z(#—1)—80. 


Remove the parentheses, 
v’— Tx+12=2?—2—30. 
Since 2 on the left and x? on the right are precisely the 
same, including the sign, they may be cancelled. 
Then —72+12=—2—8380. 
Transpose — x to the left side, and + 12 to the right side, 
—Tate=— 30-12. 
Combine, 
—6x=— 42. 
Divide by —6, x2 = 7. 
VERIFICATION. 
(7 —3)(7—4)=7(7—1)— 30, 
4x3=7x6—830, 
12 = 42 — 30, 
12 = 12 


Exercise 17. 
Find what number x stands for 
If « —5 stands for 7. 
If «+8 stands for 12. 
If 6x—12 stands for 18. . If d¢e¢—2=382+4. 
If 72— 8 stands for 25. . If 7¢a¢—5=62—1. 
If 52+8 stands for 43. 10. If 52%—8= 25 — 22. 


. If2e¢—-—5=7+2. 
. If 2e%7—4=5—2. 


i 
CO Comet | C5 


SIMPLE EQUATIONS. 65 
11. If 32 and 2+ 8 stand for the same number. 
12. If 22 —5 stands for the same number as 32. 


Solve the equations : 


13. 2a—38=8+2. 16. 84—4=12—2. 

14. 54+4=20+4 2. 17. 22—5=7—22. 

15. 22 -—3=7—~72. 18. 82+14=2—2. 
Find x 


19. If 22—5 and 4x — 11 stand for the same number. 
20. If x(x—7) and x?-- 70 stand for the same number. 
21. If x(8x—2) and 82(a—1)+2 stand for the same 


number. 


22. If 832—5=42—10. 

23. If 2a—4=14—xz2. 

24. If 3x—8 and 42 —11 stand for the same number. 
25. If 22—5 and 7 —~2 stand for the same number. 

26. If 22?— 23 and (2x+1)(x— 3) stand for the same 


number. 

27. If (~+ 38) (x—7)—(#—4) («+ 1)=25. 
28. If (22 —1)(#+3)—(a#— 38) (22 — 3) = 72. 

Solve the equations: 
29. 2(*—5)=2’— 380. 
30. r(7#+3)=2°+ 18. 
31. («—3)(@+1]l)=2—82+1. 
32. (cx—1)(+2)+ (+38) (@—-1)=22(x4+4)—(#+1). 
33. 2(x+8)—(2#+1)(a#—2)—5(#+3)+3=0. 
34. (w—3)(x+3)—(x— 4) (a+ 4)-—2=0. 


66 SCHOOL ALGEBRA. 


97, Statement and Solution of Problems. The difficulties 
which the beginner usually meets in stating problems will 
be quickly overcome if he will observe the following direc- 
tions: 

Study the problem until you clearly understand its mean- 
ing and just what is required to be found. 

Remember that z must not be put for money, length, 
time, weight, etc., but for the required nwmber of specified 
units of money, length, time, weight, etc. 

Iixpress each statement carefully in algebraic language, 
and write out in full just what each expression stands for. 

Do not attempt to form the equation until all the state- 
ments are made in symbols. 

We will illustrate by examples: 


(1) John has three times as many oranges as James, and 
they together have 32. How many has each? 


Let x be the number of oranges James has ; 
then 3a is the number of oranges John has; 
and x +3. is the number of oranges they together have. 


But 32 is the number of oranges they together have; 


“2 pOx = 32% 

or, 4a = 32, 
and t= 8, 

Since xz = 8, 3a = 24. 


Therefore James has 8 oranges, and John has 24 oranges. 
Nore. Beginners in stating the preceding problem generally write: 
Let « = what James had. 


Now, we know what James had. He had oranges, and we are to 
discover simply the number of oranges he had. 


(2) James and John together have $24, and James has 
$8 more than John. How many dollars has each? 


SIMPLE EQUATIONS. 67 


Let x be the number of dollars John has; 
then x + 8 is the number of dollars James has; 
and «# +(# + 8) is the number of dollars they both have. 


But 24 is the number of dollars they both have; 
“. @ + (x + 8) = 24: 
Removing the parenthesis, we have 
e+xe+8 = 24. 
Transposing and collecting like terms, we have 
2x2 = 16. 
Dividing by 2, we get 
x = 8, 


Since a = 8, x2+8=16, 
Therefore John has $8, and James has $16. 


Notr. The beginner must avoid the mistake of writing 
Let « = John’s money. 


We are required to find the number of dollars John has, and there- 
fore x must represent this required number. 


(3) The sum of two numbers is 18, and three times the 
greater number exceeds four times the less by 5. Find the 
numbers. 


Let « = the greater number. 
Then, since 18 is the sum, and z is one of the numbers, the other 
number must be the sum minus x. Hence 


18 ~ x = the smaller number. 


Now, three times the greater number is 3a, and four times the less 
number is 4(18 — x), 

We know from the problem that 3a exceeds 4(18 — x) by 5; or, 
in other words, we know that the excess of 3x over 4(18 — x) equals 
5. It only remains to determine what sign the word “ excess” im- 
plies. If we are in doubt about it, we can apply the phrase to two 
arithmetical numbers. We shall have no difficulty in seeing that the 
excess of 50 over 40 is 10; that is, 50 —40, and hence that the sign 
— is implied by the word “ excess.” 


68 SCHOOL ALGEBRA. 


Hence, 3a —4(18 — x) = the excess. 
But a 5 = the excess. 
“ 3a—4(18 —2) =5, 
or 3e0—72+4a=5. 
“. 1e = 77, 
and y= 11, 


Therefore the numbers are ll and 7. 


(4) Find a number whose treble exceeds 40 by as much 
as its double falls short of 35. 


Let x = the required number ; 
then 3a = its treble, 
and 3a — 40 = the excess of its treble over 40; 
also, 35 — 2” = the number its double lacks of 35. 
Hence, 32 — 40 = 35 — 2a. 
Transposing, 3a +22 = 354 40, 
“. 52 = 75, 
and % = 15, 


Therefore the number required is 15. 


(5) Find a number that exceeds 50 by 10 more than it 
falls short of 80. 


Let « = the required number; 
then x — 50 = its excess over 50, 
and 80 — a =the number it lacks of 80. 
Hence, x — 50 — (80 — x) = the excess. 
But 10 = the excess. 
“, « — 50 — (80 — x) = 10, 
or x—50—804+2=10. 
“. 22 = 140, 
and x = 70, 


Therefore the number required is 70. 


SIMPLE EQUATIONS. 69 


Exercise 18. 


1. If a number is multiplied by 7, the product is 301. 
Find the number. 


2. The sum of two numbers is 48, and the greater is five 
times the less. Find the numbers. 


3. The sum of two numbers is 25, and seven times the’ 
less exceeds three times the greater by 35. Find the num- 
bers. 


4. Divide 20 in two parts such that four times the 
greater exceeds three times the less by 17. 


5. Divide 23 into two parts such that the sum of twice 
the greater part and three times the less part is 57. 


6. Divide 19 into two parts such that the greater part ex- 
ceeds twice the less part by 1 less than twice the less part. 


7. A tree 84 feet high was broken so that the part 
broken off was five times the length of the part left stand- 
ing. Required the length of each part. 


8. Four times the smaller of two numbers is three times 
the greater, and their sum is 63. Find the numbers. 


9. A farmer sold a sheep, a cow, and a horse for $216. 
He sold the cow for seven times as much as the sheep, and 
the horse for four times as much as the cow. How much 


did he get for each? 


10. Distribute $15 among Thomas, Richard, and Henry 
so that Thomas and Richard shall each have twice as much 
as Henry. 


11. Three men, A, B, and C, pay $1000 taxes. B pays 
four times as much as A, and C pays as much as A and B 
together. How much does each pay ? 


70 SCHOOL ALGEBRA. 


12. John’s age is three times the age of James, and 
their ages together are 16 years. What is the age of each? 


13. Twice a certain number increased by 8 is 40. Find 
the number. 


14. Three times a certain number is 46 more than the 
number itself. Find the number. : 


15. One number is four times as large as another. If I 
take the smaller from 12 and the greater from 21, the 
remainders are equal. What are the numbers? 


16. The joint ages of a father and son are 70 years. If 
the age of the son were doubled, he would be 4 years 
younger than his father. What is the age of each? 


17. A man has 6 sons, each 4 years older than the next 
younger. The eldest is three times as old as the youngest. 
What is the age of each? 


18. Add $24 to a certain amount, and the sum will be 
as much above $80 as the amount is below $80. What is 
the amount ? 


19. Thirty yards of cloth and 40 yards of silk together 
cost $330; and the silk costs twice as much per yard as 
the cloth. How much does each cost per yard? 


20. Find the number whose double diminished by 24 
exceeds 80 by as much as the number itself is less than 100. 


21. In a company of 180 persons composed of men, 
women, and children there are twice as many men as 
women, and three times as many women as children. 
How many are there of each? 


22. A banker was asked to pay $56 in five-dollar and 
two-dollar bills in such a manner as to pay the same num- 
ber of each kind of bills) How many bills of each kind 
must he pay ? 


SIMPLE EQUATIONS. 71 


23. How can $3.60 be paid in quarters and ten-cent 
pieces so as to pay twice as many ten-cent pieces as 
quarters ? 


24. I have $1.98 in ten-cent pieces and three-cent pieces, 
and have four times as many three-cent pieces as ten-cent 
pieces. How many have I of each? 


Nore. In problems involving quantities of the same kind ex- 
pressed in different units, we must be careful to reduce all the quanti- 
ties to the same unit. ; 


25. I have $17 dollars in two-dollar bills and twenty- 
_five-cent pieces, and have twice as many bills as coins. 
How many have I of each? 


26. I have $6.50 in silver dollars and ten-cent pieces, 
and I have 20 coins in all. How many have I of each? 


27. A bought 9 dozen oranges for $2.00. Fora part he 
paid 20 cents per dozen; for the remainder he paid 25 cents 
a dozen. How many dozen of each kind did he buy ? 


28. A gentleman gave some children 10 cents apiece, 
and found that he had just 50 cents left. If he had had 
another half-dollar, he might have given each of them at 
first 20 cents instead of 10 cents. How many children 
were there? 


29. A is twice as old as B and 6 years younger than OC. 
The sum of the ages of A, B, and C is 96 years. What is 
the age of B? 


30. Divide a line 24 inches long into two parts such 
that the one part shall be 6 inches longer than the other. 


31. Two trains travelling, one at 25 and the other at 30 
miles an hour, start at the same time from two places 220 
miles apart, and move toward each other. In how many 
hours will the trains meet? 


12, SCHOOL ALGEBRA. 


32. A man bought twelve yards of velvet, and if he had 
bought 1 yard less for the same money, each yard would 
have cost $1 more. What did the velvet cost a yard? 


33. A and B have together $8; A andC,$10; Band C, 
$12. How much has each? 


34. Twelve persons subscribed for a new boat, but two 
being unable to pay, each of the others had to pay $4 more 
than his share. Find the cost of the boat. 


35. A man was hired for 26 days on condition that for 
every day he worked he was to receive $3, and for every day 
he was idle he was to pay $1 for his board. At the end of 
the time he received $58. How many days did he work? 


36. A man walking 4 miles an hour starts 2 hours after 
another person who walks 3 miles an hour. How many 
miles must the first man walk to overtake the second ? 


37. A man swimming in a river which runs 1 mile an 
hour finds that it takes him three times as long to swim a 
mile up the river as it does to swim the same distance down. 
Find his rate of swimming in still water. 


38. At an election there were two candidates, and 2644 
votes were cast. The successful candidate had a majority 
of 140. How many votes were cast for each? 


39. Two persons start from towns 55 miles apart and 
walk toward each other. One walks at the rate of 4 miles 
an hour, but stops 2 hours on the way; the other walks at 
the rate of 8 miles an hour. How many miles will each 
have travelled when they meet? . 

40. A had twice as much money as B; but if A gives B 
$10, B will have three times as much as A. How much 
has each ? 


41. If 22 —8 stands for 20, for what number will 4—2 
stand ? 


SIMPLE EQUATIONS. 73 


42. A vessel containing 100 gallons was emptied in 10 
minutes by two pipes running one at atime. The first pipe 
discharged 14 gallons a minute, and the second 9 gallons a 
minute. How many minutes did each pipe run? 


43. A man has 8 hours for an excursion. How far can 
he ride out in a carriage which goes at the rate of 9 miles 
an hour so as to return in time, walking at the rate of 3 
miles an hour? 


44. If 32—4a—22x-—a, find the number for which 
42 — Ta stands. 

45. If 7x—a=9(«#—a), find the number fois Be 

zta 

Norr. When we compare the ages of two persons at a given time, 
and also a number of years after or before the given time, we must 
remember that both persons will be so many years older or younger. 
Thus, if a man is now 22 years old and his son z years old, 5 years 
ago the father was 22—5 and the son x—5, and 5 years hence the 
father will be 2% + 5 and the son a + 5, years old. 


46. A man is now twice as old as his son; 15 years ago 
he was three times as old as his son. Find the age of each. 


47. A man was four times as old as his son 7 years ago, 
and will be only twice as old as his son 7 years hence. Find 
the age of each. 


48. A, who is 25 years older than B, is 5 years more 
than twice as old as B. Find the age of each. | 


49. A man is 25 years older than his son; 10 years ago 
he was six times as old as his son. Find the age of each. 


50. The difference in the squares of two consecutive 
numbers is 19. Find the numbers. 


51. The difference in the squares of two successive odd 
numbers is 40. Find the numbers. 


CHAPTER VL 
MULTIPLICATION AND DIVISION. 
SPECIAL RULES. 


98. Special Rules of Multiplication. Some results of mul- 
tiplication are of so great utility in shortening algebraic 
work that they should be carefully noticed and remem- 
bered. The following are important : 


99. Square of the Sum of Two Numbers. 
(a +8) =(a+b)(a+0) 
=a(a+ 6)+b(a+ 5) 
=@+ab+ab+ 0° 
=a?+2ab-+ 6’. 
Since a and 6 stand for any two numbers, we have 


Rute 1. The square of the sum of two numbers is the 


_ sum of ther squares plus twice ther product. 


100. Square of the Difference of Two Numbers. 
(a—b)' = (a—b)(a—8) 
= a(a—b)— b(a—d) 
=a—ab—ab+ 6 
= a? — 2ab+ 8’. 


~Hence we have 


Rue 2. The square of the difference of two numbers 1s 
the sum of ther squares minus twice ther product. 


a 


SPECIAL RULES OF MULTIPLICATION. 0 


101. Product of the Sum and Difference of Two Numbers. 
(a+ 6)(a— 6) =a(a— b)+b(a—6) 
=@—ab+tab—6 
=a — 6. 
Hence, we have 


Rue 38. The product of the sum and difference of two 
numbers rs the difference of their squares. 


If we put 2x for a and 3 for 6, we have 
Rule 1, (22+ 38/?=42?+122+4+ 9. 
Rule 2, (22— 3 =42?—1244 9. 
Rule 3, (2x +3) (2% — 3) = 427 —9. 


Exercise 19. 


Write the product of 


1. (oy) 1. (@ty)(e—y), 

2. (e—a)’. 8. (4z2—3)(4z+ 3). 

3. (7+ 26)’. 9. (807+ 407) (38¢ — 40’). 
4. (8% — 2c)’. 10. (83a—c)(8a—¢). 

5. (4y—5)’. 11. (c+ 70’) (x+ 70’). 

6. (8a? + 427)”. 12. (ax+2by) (ax — 2 by). 


102. If we are required to multiply a+6-+e¢ by a+b—e, 
we may abridge the ordinary process as follows : 
(atb+0)(a+b—c)=[(a-+b)+e][(a+b)—] 
By Rule 8, = (a+ b?—¢ 
By Rule 1, == q® + 2ab + b? — 0’, 


76 SCHOOL ALGEBRA. 


If we are required to multiply a+ 6—c by a—b-+e, we 
may put the expressions in the following forms, and per- 
form the operation : 


(a+b—0)(a—b +e) =[a+ (6-0) [a—(b—o)] 
By Rule 8, =a —(b—c/y 
By Rule 2, = o — (6° — 2bc +c’) 

=a’ — b?+ 2be— e’. 


Exercise 20. 

Find the product of 

1. 2+y+2 and 7x—y—2. 
x—y+z2 and «—y—z. 
ax -+by+1 and ax-+ by—1. 
l+a—y and 1—z+¥. 
a+ 26—8e and a—26+4+3e. 
a —ab+b* and @+ab+6". 
m+t+mn+n and m’—mn+n’. 
2+2+2? and 2—2—2". 
@V+tat+l1 and @—atl. 
da+2y—z2 and 8x—2y+z. 


CM XH xT Pw W 


a 
= 


108. Square of any Polynomial. If we put ~z for a, and 
y +2 for 4, in the identity 


(a+ 6)? =a? + 2ab +B’, 
we shall have 
[w+ (y+z)P=2? + 20(y+z)+ (y+2), 
or (a@+y+zP =2?+2ay+2a2+ 74 2y2+2 
=P LY+L + Qry + Qae+ Qyz. 


SPECIAL RULES OF MULTIPLICATION. it’ 


It will be seen that the complete product consists of the 
sum of the squares of the terms of the given expression and 
twice the products of each term into all the terms that fol- 
low it. 

Again, if we put a—6 for a, and e—d for 3, in the same 
identity, we shall have 


[(a—b) + (e—d)} 
= (a—b)’+2(a—6b)(e—d)+(e—dy 
= (a —2ab+b)+2a(e— d)—2b(e—d)+(e?—2cd+d’) 
= @—2ab+b2+2ac—2ad—2bce+2bd+ce— 2cd+d? 
= 7+674+-¢4+ d?—2ab+2ac—2ad—2be+2bd—2cd. 


Here the same law holds as before, the sign of each 
double product being ++ or —, according as the factors com- 
posing it have like or unlike signs. The same is true for 
any polynomial. Hence we have the following rule: 


Rue 4. The square of a polynomial is the sum of the 
squares of the several terms and twice the products obtained 
by multiplying each term into all the terms that follow tt. 


Exercise 21. 


Write the square of 


1.22 —3y. 10. 2+ y'+ 2’. 

2. atb+e. 11. 2a—y—z. 

3. @+y—Zz. 12. a—2b—3e. 

4. «—y+z. _ 13. 8a—6+42e. 
6. oty+sd. 14. x+2y—3z. 

6. ++2y+8. 15. «+y+2-+1. 

7. a—b+e. ; 16. 4¢+y+2-—2. 

8. 82—2y+4. 1. 2e—y—2—8. 

9. 24—8y-+ 4z. 18. «—2y—324+4. 


78 SCHOOL ALGEBRA. 


104. Product of Two Binomials. The product of two bino- 
mials which have the form x-+-a, «+4, should be carefully 
noticed and remembered. 


(1) (@+5)(e#+3)=2 («+ 38)+5 (4+ 3) 
=2+382152+15 
= 27+ 82+ 15. 


(2) (e—5)(e—8)=«(e—8)—5 (e—8) 
= 27 —82—527+15 
= z*—824+165. 


(3) (¢+5)(¢—38)=2(4—3)+5(e¢—8) 
= 2?7—32+5x—-15 
= 2712x2—15. 


(4) (*#—5)(«+3)=—2(¢+4+3)—5(e+3) 
=277+382—5x2—15 
= 2? —2a2—15. 


Each of these results has three terms. 

The first term of each result is the product of the first 
terms of the binomials. 

The last term of each result is the product of the second 
terms of the binomials. 

The middle term of each result has for a coefficient the 
algebraic swum of the second terms of the binomials. 


The intermediate step given above may be omitted, and 
the products written at once by wspection. Thus, 


(1) Multiply «+8 by «+7. 
8+7=15, 8x 7=56. 
(@+8)(@+ 7) =2?+ 15a + 56. 


SPECIAL RULES OF MULTIPLICATION. 79 


(2) Multiply «— 8 by x— 7. 


(-8)+(—1) =-15, (-8)(-7) =+86. 
. (a — 8) (x — 7) = 2? —152x-+ 56. 


(3) Multiply «—7y by x+ 6y. 


ee SS SS SS 
ao FF WO WO KF OO 


6 1. 


(—Ty) xX 6By =— 427’. 
J. (@-—- Ty) (a+ by) = 2? — ay— 42y’. 
(4) Multiply 27+ 6(a+ 6) by 2? —5(a+0). 
+6—5=1, 6(a+6)x—5(a+6)=—30(a+6)*. 
”. [2°+6 (a+0) | [2’—5 (a+6)] = a*++(a+b)2?—380 (a+)'. 


Exercise 22. 


Find by inspection the product of 


- («+8)(%+3). 
. (c+ 8)(x — 8). 
. (xc«—7)(x+ 10). 
(x — 9) (x — 5). 
» (c—10)(%+ 9). 
(a —10)(a—5). 


. (a—12)(a—8). 


- (a+ 26)(a+40). 
- (a—38b)(a+76). 
7 (a+ 26)(a— 96). 
. («—8a)(x4—4a). 


- («+ 42)(4%— 22). 


. (e+ by) (z— By). 


. («—38a)(a-+ 2a). 
- (a+ 26)(a—40). 


. (x? — 9) (2 + 8). 
(+ 2y") (a —8y). 

(at +8y") (2 4y). 

. (ab —8)(ab+5). 

. (ab —Txy)(ab+8zy). 
. (v—8y)(v—8y). 

. (x+6)(%-+ 6). 

. (a—8b)(a—385). 

. («—e)(e— ad). 

. («+ a)(x—5). 

. (c—a)(«+ 8). 

. [((a+6)+2][(a+6)—4]. 
- [(e@+y)—2][@t+y)+ 4]. 
- (@ty—T)(e+y+10). 
- (e—y—T)(«#—y—10). 


80 SCHOOL ALGEBRA. 


105. In like manner the product of any two binomials 
may be written. 


(1) Multiply 2a—6 by 3a+40. 
; (2a— b) (3a+4b) =6a?+4+ 8 ab —8ab —42? 
= 6a’?+ 5ab — 406. 
(2) Multiply 22+ 3y by 3a—2y. 
The middle term is 
(2x2) x (—2y)+3yxX 382=52y; 
“(22+ 8y)(82— 2y) = 62? + zy — 67%. 


Exercise 23. 
Find the product of 
38x2—y and 2x%+ y. 10z2—3y and 10x— Ty. 
3a?— 26? and 2a’?+ 36’. 
at? and a—Qb. 
a—Ty and 2%—5y. 3a’?— 20? and 2a+ 30. 
llz—2y and 7x+y. 10. a—O anda+8. 


4x2—8y and 3x—2y. 
5a2—4y and 8a—4y. 


o FP WO WD 
£9: Otay 


106. Special Rules of Division. Some results in division 
are so important in abridging algebraic work that they 
should be carefully noticed and remembered. 


107. Difference of Two Squares. 

From §101, (a+ 6)(a— 6)=a@ — B’. 
at 
Wasted 


Rue 1. The difference of the squares of two numbers 1s 
divisible by the sum, and by the difference, of the numbers. 





=a — 6, and Cath, Hence 
a- 


Write by inspection the quotient of 


1. 


12. 


cave 


SPECIAL RULES OF DIVISION. 


ate 5 
a—2 
ea 
3s+2 
loo 
4+a 
a — 25. 
x—5d 
SO — a 
6+ 2 


oa — sf’ 

















fn}. 


252° — 360° 
5a+6b 


49 @ — d? 


it—a 
9a?—1 





Ba+l1 


16—4@ 
4—2a 





Exercise 24. 


14. 


15. 


16. 


LT. 


18. 


19. 


20. 


aie 


22. 


23. 


24. 


25. 


abe — x? 
ab®ct + 28 
xta® — bY 
at — BE 
et oye 
a—(b+c¢) 
a Lire ed be 
a—(38b—4c) 
Dy (ed 
1+(@—y) 
(82—y)'— 162 
(82—y)+4z 
(2+3a) — Qa? 
(7 -+- 3.a)— 32 
li (la pe) 
1+ (7a—56b) 
(82+ 2yf—42* 
(82+ 2y)—2z 


(ae7 Oo Gn-V/)" 


(ap) (e—4¥) 
a (ya) 
t—(y +2) 

(1 —2y)y— 25 
(c—2y)+5 


(22 +y)? — 92 
(22+y)—382 


81 


82, SCHOOL ALGEBRA. 


108. Sum and Difference of Two Cubes. By performing 
the paren, we find that 


a + 53 
Sti es ah b?, 
Tse =a?—ab+ 


Hence, 


== at + ab + Bt 








Rue 2. Zhe sum of the cubes of two numbers ts divisible 
by the sum of the numbers, and the quotient is the sum of 
the squares of the numbers minus ther product. 


Rute 8. The difference of the cubes of two numbers is 
divisible by the difference of the numbers, and the quotient 
as the sum of the squares of the numbers plus ther product. 


Exercise 25. 


Write by inspection the quotient of 

















y Lae of OR — 
1—2z ab —e 
Palais : igp eee 
1+ 22 ab+e 
3. Bhs ei 64-Pe 
38a—b 4+y 
ae 27 at + OF is 348 — 8a 
38a+b 7—2a 
eM ee Nee 49. Oo eee 
4x+3y 2a-+ 6° 
< bara ta xf 729° 
42 —3y z+ Oy 
7. LS ces 15. at ee 
1—8z a — 3b 
et 1-272 16. 82° — 64y 





1432 2a — Ay? 


SPECIAL RULES OF DIVISION. 83 


109. Sum and Difference. of any Two Like Powers. By 
performing the division, we find that 


een 











7 =a7t@btab’?+ bd’; 
a— 
oat — ab + al? — 
a 
a’ — 5 x 
; =at+@bt+al’+ab?+ bd; 
Q— 
ote =at—ad + ab? — ab’ + bt. 
ar 


We find by trial that 
at 0, at+ bt, a®+ 0°, and so on, 
are not divisible by a+ 6 or bya—b. Hence, 


If n is any positive integer, 


(1) a®+0" ts divisible by a+b if n is odd, and by neither 
a+b nor a—b if nts even. 

(2) a®—b" is divisible by a— b if n ws odd, and by both 
a+b and a—b if ns even. 

Nore. It is important to notice in the above examples that the 
terms of the quotient are all positive when the divisor is a —6, and 
alternately positive and negative when the divisor is a + 6; also, that 


the quotient is homogeneous, the exponent of a decreasing and of b 
increasing by 1 for each successive term. 


Exercise 26. 


Find the quotient of 











iy, ane hy Aaa foe 
t—Yy x+1 x+2 

2. wy 5. vt — 16 8. Lah 
x+y e—2 l—m 
4 a 

e Hegre af = 32 9 Th om 





6. . 
x—1 x—2 l+m 


CHAPTER VIL. 
FACTORS. 


110. Rational Expressions. An expression is rational when 
none of its terms contain square or other roots. 


111. Factors of Rational and Integral Expressions. By fac- 
tors of a given integral number in Arithmetic we mean 
integral numbers that will divide the given number with- 
out remainder. Likewise by factors of a rational and inte- 
gral expression in Algebra we mean rational and integral 
expressions that will divide the given expression without 
remainder. 


112. Factors of Monomials. The factors of a monomial 
may be found by inspection. Thus, the factors of 14a%b 
are 7,2, a, a, and 0. 


118. Factors of Polynomials. The form of a polynomial 
that can be resolved into factors often suggests the process 
of finding the factors. 


CasE I. 
114. When all the terms have a common factor. 
(1) Resolve into factors 227+ 62y. 


Since 22 is a factor of each term, we have 


2% 20 = 25 
“. 22+ 62y=22(4+ 3y). 


Hence, the required factors are 2x and «+ 3y. 


FACTORS. 85 


(2) Resolve into factors 16a*+ 4a?— 8a. 
Since 4a is a factor of each term, we have | 


16a! + 4a ~8a__16a° , 4a’ 8a 


4a 4a '4a Aa 
=4¢7+a—2. 
», 16a +4a— 8a=4a(4e?+a— 2). 


Hence the required factors are 4a and 4a?+a—2. 


Exercise 27. 


Resolve into two factors: 


Biuat — 62". 7. 3a°b — 4a’. 

2. 2a?—Aa, 8. Say? + Aaty?! 

3. 5ab— 5a’d’. 9. 38a*—92?— 62’, 

4. 30—a +a. 10. 8a’a? —4a°b 4+ 12077’. 

5B. a? + ay — xy’. 11. 8a°b’c? — 4a7b°C’. 

6. at—a’b+a’l’. 12. 15a’x — 10a*y + 5a*z. 
Case II. 


115, When the terms can be grouped so as to show a common 
factor. 


(1) Resolve into factors ac + ad + be + bd. 


ae +ad + be + bd =(ac+ad)+ (be + bd) (1) 
=a(e+d)+b(e+d) (2) 
= (a+ 6)(e+d). (3) 


Nore. Since one factor is seen in (2) to be c+ d, dividing byc+d 
we obtain the other factor, a+ 6. 


86 SCHOOL ALGEBRA. 


(2) Find the factors of ae + ad — be — bd. 
ac + ad — be — bd = (ac + ad) — (be + bd) 
=a(e+d)—b(e+d) 
=(a—b)(e+d). 
Norse. Here the signs of the last two terms, — be— dd, being put 
within a parenthesis preceded by the sign —, are changed. 
(3) Resolve into factors 32° —52*—6x-+ 10. 
82° —52?— 62+ 10 = (82° — 52”) — (6x —10) 
= 27(82—5)—2(8%—5) 
= (x? — 2)(8x%—5). 


(4) Resolve into factors 52° —15axz?— 2x -+ 3a. 
52'—15az27—x+ 38a= (52° — 15az”) — (x — 8a) 
= 52? (x —8a)—1(x#-—38a) 

= (527 —1)(x— 3a). 


(5) Resolve into factors 6y — 27 xy — 102+ 452". 
6y—27 ey—102+ 452° = (452° — 27 2*y) — (102 — 6y) 
= 92? (52—3y)—2(d4 —3y) 
= (92? — 2)(5x—38y). 
Norts. By grouping the terms thus, (6y — 27a?y) — (10% — 452°), 
we obtain for the factors, (2 — 92?) (3 y — 52). 
But (2 —92?)(3y —52) = (9a? — 2)(5%—3y), since, by the Law 
of Signs, the signs of two factors, or of any even number of factors, 
may be changed without altering the value of the product. 


Exercise 28. 
Resolve into factors : 
5. 2? + ax — bx—ab. 
6. a’ + 2y — an — ay, 
ax — cy —ay+ex. 7. 2° —ay—62r+ by. 
2ab—8ac—2by4+8cy. 8. 22°—32y+4axr—Bay. 


ax — ba + ay — by. 
ax — bx — ay + by. 


PrP OO DW 


ra 


FACTORS. 87 


9. wb —abz—ac+ ex. 
10. wbx + b’ex — acy — be’y. 
11. 32° —5y7’?— 62° + 1l0zy’. 
12. 8ar—10bu—12a+415d, 
18. 227°—32?>—42+-6. 
14. 62*+ 82° — 927— 122. 
15. azt+ bx? —ax—b. 
16. 3c2*— 2dz* — 9cz’* + 6dz. 
17. 1+152* —52z— 382’. 
18. av’+t+a@et+atse. 
19. (a+6)(c+d)—3c(a+d). 
20. (c—y)'+ 2y(%@—y). 
116. If an expression can be resolved into two equal 
factors, the expression is called a perfect square, and one of 
its equal factors is called its square root, 


Thus, 162°y? = 42°y xX 4a°y. Hence, 16x°y’ is a perfect 
square, and 42*y is its square root. 
Notre. The square root of 16 2*y? may be — 4a°y as well as + 425y, 


for —4a°y x — 4a°y = 16 a*y?; but throughout this chapter the posi- 
tive square root only will be considered. 


117. The rule for extracting the square root of a perfect 
square, when the square is a monomial, is as follows: 


Extract the square root of the coefficient, and divide the 
index of each letter by 2. 


118,. In like manner, the rule for extracting the cube 
root of a perfect cube, when the cube is a monomial, is, 

Extract the cube root of the coefficient, and divide the index 
of each letter by 8, 


88 SCHOOL ALGEBRA. 


119. By §§ 99, 100, a trinomial is a perfect square, if its 
first and last terms are perfect squares and positive, and its 
middle term is twice the product of their square roots. 
Thus, 16 a? — 24ab + 90? is a perfect square. 

The rule for extracting the square root of a perfect 
square, when the square is a trinomial, is as follows: 


Extract the square roots of the first and last terms, and 
connect these square roots by the sign of the middle term. 


Thus, if we wish to find the square root of 
16a? — 24ab+ 96?, 
we take the square roots of 16a? and 90’, which are 4a 
and 30, respectively, and connect these square roots by the 


sign of the middle term, which is —. The square root is 
therefore Te eee = 


In like manner, the square root of 


1l6a@?+ 24ab+ 967? is 4a+ 30, 


CasE III. 
120, When a trinomial is a perfect square. 
(1) Resolve into factors 2’ + 22y +7’. 
From § 119, the factors of 2? + 22y+ 7’ are 
(@ry@ty). 


(2) Resolve into factors a* — 22°y + 7’. 
From § 119, the factors of #* — 22°y + 7? are 
(7? —y)@—y). 
Exercise 29. 
Resolve into factors : 
1. @—6ab+906". . 3. @&—4ab+48*, 
2. 407+ 4ab+ 0’. 4. v?+6xy+9y’. 


FACTORS. 89 


5. 427—12axr+ 9a’. 13. 492° — 282xy-+ 47’. 

6. a’ —10ab+ 250’. 14. 1— 206+ 10027. 

7 4¢7—4a+l1. 15. 8la’?+126ab+ 490". 
8. 497? —l4yz+ 2. ._ 16. mn? — 16mnda? + 64 a+. 
9. 2? —162-+ 64. 17. 4a?— 20axz + 252”. 
10. 92°+ 24zy+ 167’. 18. 12la?+ 198 ay-+ 817’. 
11. 16a?+ 8az + 2’. 19. @b*c® — 2ab’c’x’ + x”. 
12. 25+ 802+ 642’. 20. 49—140#?+ 100#*. 

CasE LV. 


121. When a binomial is the difference of two squares. 
(1) Resolve into factors 2? — y’. 
From § 101, («+y)(#-—y)=xv7—y’. 


Hence, the difference of two squares is the product of 
two factors, which may be found as follows : 


Take the square root of the first term and the square root 
of the second term. 

The sum of these roots will form the first factor ; 

The difference of these roots will form the second factor. 


Exercise 30. 


Resolve into factors : 


1. w—4, 6. 25 — 16a’. 11. 812’?— 4y’. 
2. 1—2’. 7. 16— 257’. 12. 64a‘ — dt. 
3. 2? — 97’. 8. ab? —1. 13. mn? — 36. 
4. 4a?— 490’. 9. x?— 100. 14. xt — 144. 
5. av? — 4y’. 10. 121a’—360’, 15. 2? — 25. 


90 SCHOOL ALGEBRA. 


16. 49—10077. 23. 49a*—y”. 30. 25 — 647’. 

17. 1— 492°. 24. 64a@— 90°. 31, 1621 = aye 
18. 4—121y’. 25. 8la‘dt‘—ct. 32. 252%—l6a*e’. 
19. 1—169a%. 26. 4a°’c—Q9ec’. 33. 36 a72’?—49a'. 
20. v7h?— 4c. 27. 200°b? —5ab. 34. x? —167/’. 

21. 92°— a’. 28. 8a0— 120%. 35. 1—4002*. 

22, 42%—y”. 29. 9a’?— 816". 36. 4a?c— 9c’. 


122. If the squares are compound expressions, the same 
method may be employed. 


(1) Resolve into factors (#-+ 3y) — 16a’. 


The square root of the first term is x + 3y. 

The square root of the second term is 4a. 

The sum of these roots is +3y + 4a. 

The difference of these roots is a + 3y —4a. 

Therefore (x + 3y)?— 16a?=(a+3y + 4a)(a+3y—4a). 


(2) Resolve into factors a? — (36 — 5c)’. 


The square roots of the terms are a and (3b — 5c). 

The sum of these roots is a + (36 — 5c), ora +36 —5e. 

The difference of these roots is a — (36 — 5c), ora—3b+5e. 
Therefore a? — (3b —5c)l?=(a+3b—5c)(a—3b + 5e), 


123. If the factors contain like terms, these terms should 
be collected so as to present the results in the simplest 
form. 


(3) Resolve into factors (8a-+ 56)? — (2a — 3b)’. 


The square roots of the terms are 3a +56 and 2a—3b, 
The sum of these roots is (3a + 5b) +(2a— 3b), 
or 8a+5b+ 2a—3b=5a+ 2b. 
The difference of these roots is (3a + 5b) ~(2a— 86), 
or 3a+65b—2a+3b=a4+4 86, 
Therefore (3a + 5b)? ~ (2a —3 bd)? = (5a + 2b) (a + 85). 


FACTORS. 91 


Exerc:se 31. 


Resolve into factors : 





1. (¢+y)— 2. 11. (a— 6b)’ —(e—dy. 

2. («—yy—2’. 12. (2a+ by — 25e’. 

3. (x—2yy—42. 13. («+ 2y)— (Qa —y). 

4. (a+ 36) — 16’. 14. (x +38)—(382—4). 

5. wv? — (y—2)’. 15. (a+6—c)—(a—b—c)’. 
6. a’ — (86 — 2c)’. 16. (a—38x)?— (8a— 22). 

7. D—(2a+ 3c)’. 17. (2a—1)—(8a+1)’. 

8. 1—(#+ 50)’. 18. (x—5)?— (a+ y—5)*. 

9. 9a?—(x— 3c) 19. (2a+6—c)—(a—2b+e)*. 
10. l6a@’—(2y—382z). 20. (a+26—8c)?—(a+5e)’. 


124, By properly grouping the terms, compound expres- 
sions may often be written as the difference of two squares, 
and the factors readily found. 

(1) Resolve into factors a? —2ab+ b? — 9c’. 

a —2ab4+ 0-9? =(v—2ab+ 0’)—9¢ 
=(a—byP—9¢? 
=(a—b+3c)(a—b—3c). 

(2) Resolve into factors 12ab+ 92°—4a’— 90’. 


Norse. Here 12 ab shows that it is the middle term of the expres- 
sion which has in its first and last terms a? and b?, and the minus 
sign before 4a? and 90? shows that these terms must be put in a 
parenthesis with the minus sign before it, in order that they may be 
made positive. 


The arrangement will be 
92’ — (4a? —12ab+ 9b’) =92?—(2a—36)’ 
= (82+ 2a—36)(8%—2a+30). 


92 


SCHOOL ALGEBRA. 


(3) Resolve into factors — a?+ 6?—c’+d?+ 2ac-+ 26d. 


Nort. Here 2ac, 2bd, and —a?, —c?, indicate the arrangement 


required. 


—av+b—e?+d?+2ac4+ 2bd 


= (6? + 2bd+ d’) — (@ —2ac+c’) 
= (b+ d)' — (a—oF 
= Gidto jet ds ia 


Exercise 32. 


Resolve into factors : 


me © tO 


9. 
10. 
ii: 
12. 
13. 
14. 
15. 
16. 


vt+2ab+ ?— 

x’ —2ry+y'— 9a’. 
6? — 27+ 4ar —4a’. 
4¢° + 4ab+ 8? — 2”. 


av—xz—y—22y. 
1—a@’— 2ab— 86’. 
av + 6? + 2ab — 16076’. 
4¢7—9a+ 6a—-1. 
a+b? — ?— d’?—2ab — 2cd. 

w+ y? — 2xy — 2ab— a’ — B’. 
927—6x2+1—a’—4ab — 467. 

a + 2ab — 2? — bay — 97+ 8’. 

2 —2¢4+1—0?+ 2by—y’. 
9—62+27— a — 8ab— 1687. 
4—42+2°—4a—1—4¢d. 

a’ — a? —9+6'+ 6a—2a’8’. 


OOne te Com Ot 


125, A trinomial in the form of a‘t+a’7b?+6' can be 
written as the difference of two squares. 

Since a trinomial is a perfect square when the middle 
term is twice the product of the square roots of the first 
and last terms, it is obvious that we must add ad? to the 
middle term of a*-+ a’b?+ b* to make it a perfect square. 


FACTORS. 93 


We must also subtract ab? to keep the value of the 
expression unchanged. We shall then have 


(1) at+ a7? + Of = at+ 2070? + bt — a7? 
= (a+ 0) — ab? 
= (a+ 0? + ab)(a@ + 0? — ab) 
= (a + ab + 8’) (av — ab + 0’). 
If in the above expression we put 1 for 6, we shall have 
(2) @+e@+4+1=(e+2e+1)-e 
3 (a? of . 1) ta fo 
=(¢7+1+a)(7?+1-a) 
=(v7+a+I1)(?—a+l). 


(3) Resolve into factors 4a* — 37 a°y’ + 97. 


Twice the product of the square roots of 4a* and 9y* is 122747. 
We may separate the term —37.27y? into two terms, —12a?y? and 
— 25 2x?y?, and write the expression 


(4a* — 12277? + 9y*) — 25277’ 
| = (2277 — 37’) — 2527 
= (27 —3y+5xy) (20-37 —52y) 
= (227?+52y—37")(22°—odoay—3y). 


Exercise 33. 


Resolve into factors : 


1. at+ay+y. 6. 9a‘t+ 26 a7b? + 25 *. 
2. ata’+l. 7. 424*— 21277’ + 97. 
3. 9at—15a7?+1. 8. 4at— 29 a7%c? + 25c’. 
4. 1l6a*—17a@’+1. 9. 4a*+ 16a’? + 25ct*. 
5. 4a*—138¢°+1. 10. 25a*-+ 312°’? + 16y*. 


94 SCHOOL ALGEBRA. 


CasE V. 
126. When a trinomial has the form x’?+ ax-+-b. . 


From § 107 it is seen that a trinomial is often the product 
of two binomials. Conversely, a trinomial may, in certain 
cases, be resolved into two binomial factors. 


127. If a trinomial of the form 2?+ ax+6 is such an 
expression that it can be resolved into two binomial fac- 
tors, it is obvious that the first term of each factor will be 
xz, and that the second terms of the factors will be two 
numbers whose product is 4, the last term of the trinomial, 
and whose algebraic sum is a, the coefficient of x in _the 
middle term of the trinomial. 


(1) Resolve into factors «? + lla + 380. 


We are required to find two numbers whose product is 30 and 
whose sum is 11. 

Two numbers whose product is 30 are 1 and 30, 2 and 15, 3 and 
10, 5 and 6; and the sum of the last two numbers is 11. Hence, 


x + 1llx+30=(#+5) («+ 6). 


(2) Resolve into factors 2? — 72+ 12. 


We are required to find two numbers whose product is 12 and 
whose algebraic sum is — 7. 

Since the product is + 12, the two numbers are both positive or both 
negative; and since their sum is — 7, they must both be negative. 

Two negative numbers whose product is 12 are — 12 and —1, —6 
and —2, —4 and —3; and the sum of the last two numbers is —7,. 
Hence, 


2’ —T*x+12= (¢ —4)(¢#—8). 
(3) Resolve into factors 2? +4 2% — 24. 


We are required to find two numbers whose product is — 24 and 
whose algebraic sum is 2. 


FACTORS. 95 


Since the product is — 24, one of the numbers is positive and the 
other negative; and since their sum is +2, the larger number is 
positive. 

Two numbers whose product is — 24, and the larger number posi- 
tive, are 24 and —1, 12 and — 2, 8 and —3, 6 and — 4; and the sum 
of the last two numbers is +2. Hence, 


x + 2% — 24 = (x + 6) (x — 4). 


(4) Resolve into factors 2? — 3x — 18. 


We are required to find two numbers whose product is — 18 and 
whose algebraic sum is — 3. 

Since the product is — 18, one of the numbers is positive and the 
other negative; and since their sum is —3, the larger number is 
negative. 

Two numbers whose product is —18, and the larger number nega- 
tive, are —18 and 1, —9 and 2, —6 and 38; and the sum of the last 
two numbers is —3. Hence, 


x —38x—18=(x# — 6)(x+ 8). 


(5) Resolve into factors 2? —102y + 97’. 


We are required to find two expressions whose product is 9 y? and 
whose algebraic sum is —10y. 

Since the product is + 9y?, and the sum —10y, the last two terms 
must both be negative. 

Two negative expressions whose product is 9y?, are —9y and — y. 
—3y and —3y; and the sum of the first two expressions is —10y. 
Hence, 


2—=10ay+97 = (#—9y)(4#—y). 


Exercise 34, 


Resolve into factors : 


1. 2?+82+415. 4. «?—8x—10. 
2. 2 —8r+ 15. 5. 27+ 5ar-+ 6a’. 
3. 2?+ 24-15. 6. 2’?— 5axr-+ 6a’. 


96 


SCHOOL ALGEBRA. 


xv? — 2a — 15, 
zv’+5a2+6. 
x*— 52+ 6. 

. v@ta—ob. 

. v2 —a—Bb. 

. 2&+6x-+ 5, 

. 2 —62+5. 

. @+4xe—5. 

. @& —44—5., 

. 2@+9e418. 

. v2 —9x+18. 

. 2+82—18. 

. #&—dsx— 18. 

. #2 +9r+8. 

. 2&—9x+8. 

. v@t+tTe—8. 

. #@—Tx—8. 

. &@+7¢+10. 

. 2 —Tx+10. 

. 2+3a—10. 

. 2 — 5ax— 50a’. 

. vy? —382y —A4. 

. &@— Bax — 542". 

. &—ae— 2c’. 

- ff — 8yz+ 1872’. 


32. 
33. 
34. 
35. 
36. 
37. 
38. 
39. 
40. 
41. 
42. 
43. 
44. 
45. 
46. 
47. 
48. 
49. 
50. 
BL 
52. 
53. 
54. 
55. 
56. 


xv’ +ax— 6a’. 


x’ —ax— 6a’. 


e+ bay +4y’. 
x? — day —Ay’. 
x” — Say + 4y’. 
x’ + d32y—4y’. 
2 —32y —4y/’. 
a’*®— Tab + 100’. 


aa? — 8axn — 54. 

xv’ — Ta — 44. 

xv’ + 4 — 182. 

x’ —15x2+50. 

a? — 23a+ 120. 
v+17a— 890. 

c? + 25¢e— 150. 

c? — 58c-+ 57. 

a‘ — 11a’? + 308°. 
2+ Ozy + 2077. 
xy? + 19 xyz + 48 2” 
vb? — 13 abe + 22’. 
a’ — 16ab — 860". 
x +17 ay + 807’. 
x — Tay — 18y’. 

c’ + ¢ — 20. 

a’ + 16ab — 26087. 


FACTORS. 97 


57. yu? — Syz— 842’. 

58. x? —1l2— 152. 

59. 18—32—2?=— (27+ 327 — 18). 
60. 8834+ 8x2—27=— (2? — 8x — 83), 
61. 78 — Tx — 2x’. | 


Case VI. 


128, When a trinomial has the form ax?-+ bx-+ce, 
From § 105, 


(82 — 2)(5x+ 3) 
= 152°+9x2—102—6 = 152°— 2x—6: (1) 


(8x2 — 2)(52 — 8) 
= 152?—92— 102+6=152'—19246. (2) 


Consider the resulting trinomials : 
The first term in (1) and (2) is the product 3a x 52. 
The middle term in (1) is the algebraic sum of the products 


32 xX 3 and (— 2) x 5a. 

The middle term in (2) is the algebraic sum of the products 
3a xX (—3) and (—2) x 52. 

The last term in (1) is the product (— 2) x 3. 

The last term in (2) is the product (— 2) x (— 3). 


The trinomials have no monomial factor, since no one of their 
factors has a monomial factor. Hence, 


1. If the third term of a given trinomial is negative, 
the second terms of its binomial factors will have unlike 
signs. 

2. If the third term is positive, the second terms of its 
binomial factors will have the same sign, and this sign is 
the sign of the middle term. 


98 SCHOOL ALGEBRA. 


3. If a trinomial has no monomial factor, neither of its 
binomial factors can have a monomial factor. 


(1) Resolve into factors 62? + 172+ 12. 


The first terms-of the binomial factors must be either 62 and 2, or 
3a and 22. 

The second terms of the binomial factors must be 12 and 1, or 6 
and 2, or 3 and 4. 

We therefore write 


I. (624+ )(@+ J; or Il Cat” Que 


For the second terms of these factors we must reject 1 and 12; for 
12 put in the second factor of I. would make the product 6a x 12 too 
large, and put in the first factor of I., or in either factor of IL., the 
result would show a monomial factor. 

We must also reject 6 and 2; for if put in I. or II. the results 
would show monomial factors; and for the same reason we must 
reject 3 and 4 for I. 

The required factors, therefore, are (3a + 4) and (2a + 3). 


(2) Resolve into factors 142? — ll2— 15. 
For a first trial we write 


(Ter Aten e): 


Since the third term of the given trinomial is —15, the second 
terms of the binomial factors will have unlike signs, and the two 
products which together form the middle term will be one +, and 
the other —. Also, since the middle term is —1la, the negative 
product will exceed in absolute value the positive product by —1la. 

The required factors, therefore, are (77 + 5) and (2a — 3). 


Exercise 35. 
Resolve into factors : 
1. 22?°+52-+3. 4, 242’ —2ay—15y*. 
2. 827—x—2. 5. 38627?— 19ay— by’. 
3.0” —82-+ 3. 6. 1527+ 192y-+ 6y’. 


— 


FACTORS. 99 


a dan oe 14. 152° — 26xy+ 87’. 
8. 62° —4— 2. 15. 927+ 62y — 8y’. 
9. 1527+ 1427-8. 16. 62? —2y — 85y’. 
10. 82?—10x+ 3. 17. 102?— 2lzy — 107’. 
11. 1827+ 92—2. 18. 142?—55ay +217. 
12. 242*°— l4xy — 5y/’. 19. 62? — 23 zy + 207’. 


13. 240° 882y+157%. 20. 627+ 35ay—6y’. 








Casz VII. 
129. When a binomial is the sum or difference of two cubes. 
: 3 
From § 108, tae Herter 
and cag hay Mme Se 
a—b 
a +b =(a+b)(a—ab+0’) 
and a’ — B= (a — b) (a + ab + 0’). 


In like manner we can resolve into factors any expres- 
sion which can be written as the sum or the difference of 
two cubes. 

(1) Resolve into factors 8a° + 270°. 


Since by § 118, 8a* = (2a)*, and 276°= (807)*, we can 
write 8a? + 270° as (2a)? + (807). 


Since a + 6? = (a+ b)(a — ab + 0B’), 
we have, by putting 2a for a and 30? for 6, 
(2a)8+ (30°) = (2a+30*) (4a°—6ab?+ 90%). 


100 SCHOOL ALGEBRA. 


(2) Resolve into factors 1252* — 1. 
12527—1=(5z)—-1 
= (54 —1) (2527+ 52+1). 
(3) Resolve into factors 2° + 7’. 
+= (e+) 
=@ + y)e— vy ty). 


130, The same method is applicable when the cubes are 
compound expressions. 


(4) Resolve into factors (x — y)? + 2’. 
Since a@§+0?=(a+6)(?—ab+0’), 
we have, by putting «—y for a and z for 3, 
(ex—yfP+2=[(e—-y)+2][@—-yl—-@—yer2] 
= («—y+z)(a’—2ayt+y—ax2+y2+2). 


Exercise S36. 


Resolve into factors : 


1. a& + 86%. 5. 27a y® — 1. 9. 216.a° — b*. 
2. a&— 27a’. 6. a+ 270°. 10. 64a’— 276°. 
3. a+ 64. 7. vy? — 64. 11. 343 — 2°. 
4. 125a°+1. 8. 64a°-+ 1256°. 12. a’b* + 348. 
13. 8a°— J. 19. 82° —(x#—y)’. 
14. 216m* + n°. 20. 8(x+y)+ 2. 
15. (a+6)—1. 21. 7297° — 642°. 
16. (a—6b)+1. 22. (a+b) —(a— by’. 


17. (2a+y)—(x—y). 23. 729a°4+ 2164 
18. 1—(a— 6)’. 24. xy? — 5122. 


FACTORS. 101 


181. We will conclude this chapter by calling the stu- 
dent’s attention to the following statements : 


1. When a binomial has the form 2” — y”, but cannot be 
written as the difference of two perfect squares, or of two 
' perfect cubes, it is still possible to resolve it into two fac- 
tors, one of which isa—y. Thus (§ 109), 


a — 88 = (a—b) (at + a'b + a°B? + ad? + 0). 


2. When a binomial has the form 2”+ y”, but cannot be 
written as the sum of two perfect cubes, it is still possible 
to resolve it into two factors, except when 7 1s 2, 4, 8, 16, 
or some other power of 2. Thus (§ 109), 


a + 6° = (a+ b)(at—a*b + oh? — ab’ + D*). 


But a?+ 0?, at +0, a®+ 6°, cannot be resolved into 
factors. 


3. The student must be careful to select the best method 
of resolving an expression into factors. Thus, a°— 6° can 
be written as the difference of two squares, or as the dif- 
ference of two cubes, or be divided by a — 4, or by a+ 8. 
Of all these methods, the best is to write the expression as 
the difference of two squares, as follows 


(a’)? Se (0°)? == (a? + b*) (a? ac b*) 
= (a+b)(a’— ab+b’)(a—b)(a’+ab+0’). 


4, From the last example, it will be seen that an expres- 
sion can sometimes be resolved into three or more factors. 


8 eA 68 — (at + b*) ey es b*) 
— (a* + b*) ix + b) Ga Ee 6°) 
= (a + 5‘) (a? + 8’) (x + b) (a — 4). 


102 SCHOOL ALGEBRA. 


5. When a factor occurs in every term of an expression, 
this factor should first be removed. ‘Thus, 
82? —50¢0+ 42—1l0a= 2(42? — 25a? + 24 — 5a) 
= 2[ (42? — 25a”) + (22 —5a)] 
= 2(2%—5a)(2z2+5a+1). 


6. Sometimes an expression can be easily resolved if we 
replace the last term but one by two terms, one of which 
shall have for a coefficient an exact divisor or a multiple of 
the last term. ‘Thus, 

(1) #—52?+112—15=(2'—52°+62x)+(5x—15). 
=x (2*—52+6)+5(x—3) 
=(#—3)[#(@—2) +5] 
= (%—8) (x’?—2x-+5). 


(2) 2—92?+ 262—24=(2°—92?4+ 1427)+(124—24) 
= (0!—9e-+14)412("—2) 
=x (x—7)(«—2)+12(x—2) 
= (# — 2) (#’?—T2+12) 
= («—2)(x—8) (a—A4). 


(3) a — 26% — 5 = (2° — 25x) — (e+ 5) 
= x(x? — 25) — (x#+ 5) 
= («+ 5) (a — 52 —1). 
(4) a+ 3a? —4= (2+ 227)+4+ (2? —4) 
= x" (x + 2)+ (a? — 4) 
= (4 + 2) (2? + x — 2) 
= (e+2)(2+2)(e—1). 


vo 
S) 


FACTORS. 


103 


Exercise 37. 


EXAMPLES FoR REVIEW. 


Resolve into factors : 


a — 9a. 


etevtetl. 
v—2y+ 2x — xy. 


x2’ —14x%+ 49. 
36 27 — 497’. 
etna 
(c7—y)’ — 6. 
x+y’. 

. o—y, 

at -- ¥*, 

. 2 —(a—b)*. 


pti rae gies eb i aS a ee ce 


Ce ee ee — | 
o Ff oO WD FF} © 


- @—(m+n)’. 
. 2@—l1l2e+18s. 
. 2+4a— 45. 
. 2@+182+ 86. 
. #& — 184 — 48. 
. v2 +9x— 86. 
- 1027+ 2—21. 
- 62?—x— 12. 


no wo —|§ §|-& = & 
= ©Oo2 © © +t 


xy’? — 4xy* — 32°y’. 


82° + 22°—9x—6. 


. W+2mn+n?— 1. 


23. 
24. 
25. 
26. 
Bhs 
28. 
29. 
30. 
31. 
32. 
33. 
34. 
35. 
36. 
37. 
38. 
39. 
40. 
41. 
42. 
43. 
44. 


122?—xz—1, 
1227 — x — 20. 
9a¢7+ 124a+4. 

av — 6 — + 2be. 
at + aty?+ y4 

2 —6z—40. 

x’? — Tx — 60. 
a’ —19a-+ 84. 
+ 2ar+ 38b2+6ab. 
V+ mn’? —n® —2mxz. 
4Ax*— x’. 

ei ty", 

Dat + 21 x77? + 25 y'. 
w—4+4y 4+ 22y. 
227+ 3xy —2y’. 
2¢7—7Ta+6. 

x —Te+l. 
1—a’?— 0° — 2ab. 
32* — 62° + 927. 
v— 52? — 22+ 10. 
x’ t+ ax — bx — ab. 


2a? — 3xy + 4axr— bay. 


104 SCHOOL ALGEBRA. 


45. az'+ ba’ — ax— b. 63. 6a —a—T7. 
46. &+6+a+b. 64. 5c*— 15c? — 90’. 
47. &—b+a—Bb. 65. aux—cx+ay—cy. 


48. (x—y)—2y(x—y). 66. 162*—8l. 

49. 1+ 102y+ 202’°y’. 67. v+a2’+1. 

50. a—0?+ 2be—’. 68. 272° — 64a’. 

51. a? +4y7—2—4ay. 69. 2° +7. 

52. a —40?—97?+12bc. 70. 2—y’. 

53. 49°4+9y—2—122y. 71. a°— 206. 

54. (a+b) —(e—d)’. 72. 2+ l6a’2’+ 256a%. 


55. w+ y’. 73. 1—(«#—yy. 

56. 322° —c’. 74. (a@ty)4+ (Qz—y). 
57. a®+ 647%. 1b, 2216: 

BS ahi ee 176. 382° +2—2. 

bos go ay", 717. 2—8a— 2a’. 

60. (a+ 6)* —1. 78. 4—5e—6e’. 

61. v@&—B’?+a—b. 79. 2ry—2#?—y +2. 

62. v@ta+36—90’. 80. 4at— 9a@?+ 6a—1. 


81. w@ —2ab+ 0’ + l2ay —42°— 97’. 

82. 2a7—4ry + 2y’+ Zax — ay. 

83. (a+ 6)—1—ab(a+b+1). 

84. w—a’?+38x2+5. (See § 181, 6.) 

85. 62° — 2327+ 16x—38. (See § 1381, 6.) 

86. wv +y7+2—2Qxry—222+2yz. (See § 103.) 
87. 4a°b?— (a + 0? — ¢’)’. 


CHAPTER VIII. 


COMMON FACTORS AND MULTIPLES. 


132. Common Factors. A common factor of two or more 
numbers is an integral number which divides each of them 
without a remainder. 


133. A common factor of two or more expressions is 
an integral and rational expression which divides each of 
them without a remainder. Thus, 5a@ is a common fac- 
tor of 20@ and 25a; 327y? is a common factor of 1227 
and 152%’. 


134, Two numbers are said to be prime to each other 
when they have no common factor except 1. 


135, Two expressions are said to be prime to each other 
when they have no common factor except 1. 


136. The highest common factor of two or more numbers is 
the greatest number that will divide each of them without 
a remainder. 


187. The highest common factor of two or more expres- 
sions is the expression of highest degree that will divide 
each of them without a remainder. Thus, 3a’ is the highest 
common factor of 8a7, 6a’, and 12a‘; 527’ is the highest 
common factor of 102*y? and 15 2’°y’. 

For brevity, we use H.C. F. to stand for “ highest com- 
mon factor.” 


106 SCHOOL ALGEBRA. 


To find the highest common factor of two algebraic 
expressions : 


Case I. 


138. When the factors can be found by inspection. 
(1) Find the H.C. F. of 42.a°0? and 6076‘. 


A200 =2x 3X TX aaa X00" 
60a7b*= 2x23 x5 xX aa X bbDd. 
., the H.C. F.=2 x 8 x aa x 68, or 6a’d*. 


2) Find the H.C. F. of 2a?a-+ 2a2z? and 8abay+ 3 b2’y. 
y y 


2e0¢%+2an7 =2ar(a+2); 
Babry + 3bx’y = 3bry(a+2z). 
peg nate va Lik Ba thy =2(a+ 2). 


(3) Find the H.C. F. of 477+ 42 —48, 62?— 482 +4 90. 
4Avt 44—48=—4(2?+ 2-12) 
=4(x—38)(#+4 4); 
62? — 482 — 90 = 6 (2? — 8x4 4+ 15) 


= 6(e—8)(«—5); 
RDN Rats oad bed 9 Od = 2(a — 3) 
= 2a — 6. 


Hence, to find the H.C. F. of two expressions: 


Resolve each expression into its simplest factors. 

Find the product of all the common factors, taking each 
factor the least number of tumes rt occurs in any of the given 
CXPTESSLONS. 


COMMON FACTORS AND MULTIPLES. 107 


Exercise 88. 


Find the H.C.F. of 


1. 120 and 168. 4. 36a°2? and 28 2°%y. 
2. 362° and 272%. 5. 48a7b%c and 60a°c’. 
3. 42a72° and 60 a°2’. 6. 8(a+ 6) and 6(a+ 5)’. 


7. 12a(@+y) and 46(x4+ y). 
8. (7—1)?(x+ 2) and (x — 8) (x + 2)*. 
9. 2407)? (a+b) and 42a0°%b(a+ bd)’. 
10. 2’?(a—3) and 2’—3xz. 12. 2 —42 and w—62+8. 
11. a — 16 and x + 42. 18. 2? —7Tx2+12 and 2?—16. 
14. 92°— 47’ and 122’ — xy — 67’. 
15. 2 —Ta2#—8and 2’?+ 5274+4. 
16. x2’? + 32y — 107 and 2? — 22y — 35 y?. 
17. xt — 22° As? and 62°— 62 — 1802". 
18. 2° — 3a’y and 2 — 277. 
19. 1+ 642’ and 1—42-+4 162”. 
20. 2*— 81 and 2*+ 82’ — 9. 
21. #+22—3, 2+ T2+12. 
22. x?—62+5, 27+ 382 — 40. 
23. 8at+ 15a°b — 727d", 6a? — 38007) + 36.07. 
24. 62°y —122y?4+ 6y’*, 32° + 9ay’ — 127%. 
25. 1—16c*, 1+ c’—12c*. 
26. 82° +22%—1, 6a°+ 77442. 
27. 627? +a2—2, 122°—2—6. 
28. 152° —19277 + 62y’, 102*— a*y — 32°77. 
29. 102*y + 927y? —9ay*, 4a? +157°* — 42’y,. 


108 SCHOOL ALGEBRA. 


Case II. 
139. When the factors cannot be found by inspection. 


The method to be employed in this case is similar to 
that of the corresponding case in Arithmetic. And as in 
Arithmetic, pairs of continually decreasing numbers are 
obtained, which contain as a factor the H.C. F. required, 
so in Algebra, pairs of expressions of continually decreas- 
ing degrees are obtained, which contain as a factor the 


H.C. F. required. 


140, The method depends upon the following principles: 


(1) Any factor of an expression is a factor also of any 
multiple of that expression. 


Thus, if ¢ is contained 3 times in A, then ¢ is contained 
9 times in 8.4, and m times in m A. 


(2) Any common factor of two expressions is a factor of 
ther sum, thew difference, and of the sum or difference 
of any multiples of the expressions. 


Thus, if ¢ is contained 5 times in A, and 8 times in B, 
then e¢ is contained 8 times in A+B, and 2 times in 
A— B. 

Also,in5 A+2 Ait is contained 5x 5+28, or 81 times, 
and in 5A —2 28 it is contained 5X 5—28, or 19 times. 


(3) The H.C. F. of two expressions is not changed vf one 
of the expressions is diided by a factor that is not a factor 
of the other expression, or 1f one is multiphed by a factor 
that is not a factor of the other expression. 

Thus, the H.C.F. of 4a7bc? and a’c’d is not changed if 
we remove the factors 4 and 6 from 4a7bc’, and d from 


ved; or if we multiply 4a%c? by 7, and a’e*d by 11, 


. 
COMMON FACTORS AND MULTIPLES. 109 


141, We will first find the greatest common factor of 
two arithmetical numbers, and then show that the same 
method is used in finding the H.C.F. of two algebraic 
expressions. 


Find the greatest common factor of 18 and 48. 
18) 48 (2 
36 
12)18(1 
12 


6) 12(2 
12 

Since 6 is a factor of itself and of 12, it 1s, by (2), a fac- 
tor of 6+ 12, or 18. 

Since 6 is a factor of 18, it is, by (1), a factor of 2 x 18, 
or 86; and therefore, by (2), it is a factor of 86+12, or 48. 

Hence, 6 is a common factor of 18 and 48. 

Again, every common factor of 18 and 48 is, by (1), a 
factor of 218, or 36; and, by (2), a factor of 48 — 36, 
or 12. 

Every such factor, being now a common factor of 18 and 
12, is, by (2), a factor of 18 — 12, or 6. 

Therefore, the greatest common factor of 18 and 48 is 
contained in 6, and cannot be greater than 6. Hence 6, 
which has been shown to be a common factor of 18 and 48, 
is the greatest common factor of 18 and 48. 


142, It will be seen that every remainder in the course of 
the operation contains the greatest common factor sought ; 
and that this is the greatest factor common to that remain- 
der and the preceding divisor. Hence, 


The greatest common factor of any dwisor and the corre- 
sponding dividend is the greatest common factor sought, 


110 SCHOOL ALGEBRA. 


143, Let A and & stand for two algebraic expressions, 
arranged according to the descending powers of a common 
letter, the degree of B being not higher than that of A. 

Let A be divided by JB, and let Q stand for the quo- 

ent, and # for the remainder. Then 


B) A(Q 
BQ 
te 
Whence, Rk = A— BQ, and A= BQ+ BR. 


Any common factor of B and & will, by (2), be a factor 
of BQ+ R, that is, of A; and any common factor of A and 
B will, by (2), be a factor of A — BQ, that is, of A. 

Any common factor, therefore, of A and B is likewise 
a common factor of B and R&. That is, the common fac- 
tors of A and B are the same as the common factors of B 
and #; and therefore the H.C.F. of & and Z& is the 
H.C.F. of A and B. 

If, now, we take the next step in the process, and divide 
B by R, and denote the remainder by S, then the H.C. F. 
of S and A can in a similar way be shown to be the 
same as the H.C.F. of B and A, and therefore the H.C. F. 


of A and B; and so on for each successive step. Hence, 


The H.C. F. of any dwisor and the corresponding diw- 
dend 1s the H. C.F. sought. 


If at any step there is no remainder, the divisor is a fac- 
tor of the corresponding dividend, and is therefore the 
H.C.F. of itself and the corresponding dividend. Hence, 
the last disor is the H.C. F. sought. 


Notr. From the nature of division, the successive remainders are 
expressions of lower and lower degrees. Hence, unless at some step 
the division leaves no remainder, we shall at last have a remainder 
that does not contain the common letter. In this case the given 
expressions have no common factor. 


COMMON FACTORS AND MULTIPLES. PEL 


Find the H.C.F. of 227+ 2—3 and 42°+82’?—x—-6. 
207 +2—3)42°+82?7— x—6(2¢4+8 


4° + 227—6x 
627+ 52—6 
62°+ 382— 9 
22+3)22+ #¢—3(¢4—1 
227+ 32 
—2x4—8 
pees. G.b. = 22 + 3. —2x2—838 


Each division is continued until the first term of the remainder is 
of lower degree than that of the divisor. 


144, This method is of use only to determine the com- 
pound factor of the H.C.F. Simple factors of the given 
expressions must first be separated from them, and the 
H.C.F. of these must be reserved to be multiplied into the 
compound factor obtained. 


Find the H.C.F. of 
12a* + 802° — 7227 and 322°+ 842’— 1762. 
122+ + 802° — 722? = 62° (227 + 5x — 12). 
322° + 842? —176x=—42(82' + 21x — 44). 
62? and 4x have 2x common. 
Qa? + 5x2 —12)8a7+21a—44(4 
827+ 20x --48 
e+ 4)20°+52—12(2¢ —3 
227° + 82x 
—38x2—12 
». the H.C. F. = 22(x + 4). —3x—12 


112 SCHOOL ALGEBRA. 


145. Modifications of this method are sometimes needed. 


(1) Find the H.C. F. of 42?—8xz—5 and 122°—42—65. 
4a? —8x—5)122?— 4x—65(8 
122? — 24x%—15 


202 — 50 


The first division ends here, for 20a is of lower degree than 42”. 
But if 20%—50 is made the divisor, 42? will not contain 20z an 
integral number of times. 

The H.C. F. sought 7s contained in the remainder 20% — 50, and is 
a compound factor. Hence if the simple factor 10 is removed, the 
H.C. F. must still be contained in 2a — 5, and therefore the process 
may be continued with 2 — 5 for a divisor. 


2% —5)427— 8xex—5(2r+1 


427— 102 
Qu 5 
9x2—5 





.. the H.C. F. = 22 —5,., 


(2) Find the H.C.F. of 
212° — 42? — 154 — 2 and 212° — 322? — 542 — 7. 


212° — 4a? — lda— 2) 212° — 322? —54e—7(1 
QM e— 427°— 15372 


— 2827 —39%—5 


The difficulty here cannot be obviated by removing a simple factor 
from the remainder, for — 282?— 39a%—5 has no simple factor. In 
this case, the expression 21 #3 — 4a?— 15a — 2 must be multiplied by 
the simple factor 4 to make its first term exactly divisible by — 28 a”. 

The introduction of such a factor can in no way affect the H.C. F. 
sought, for 4 is not a factor of the remainder. 

The signs of all the terms of the remainder may be changed; for 
if an expression A is divisible by — F, it is divisible by + F. 

The process then is continued by changing the signs of the re- 
mainder and multiplying the divisor by 4. 


COMMON FACTORS AND MULTIPLES. 113 


2827+ 3944+ 5)842?— 1l62°?— 60x— 8(8x 
842° + 1172+ 1dex 
—1832°7— Tdx— 8 
Multiply by —4, —4 
53227 + 3002 +4 32(19 
53822? + 7412+ 95 


Divide by — 68, — 63)— 4412 — 63 
Tatil 


Tx + 1)282?+4+ 894+ 5(42+5 
2827+ 42 
85x2+5 
reine HH. O.b=/ 2+ 1. 35x2+5 


(3) Find the H.C. F. of 
827+ 2x7 —8 and 62°+ 52’ — 2. 


627+ B5a’?— 2 
these alr | 
82? + 24 —3)242°+ 202*?— 8 (82+7 
2427+ 6a°?— Ya 
1427+ 9x— 8 
Multiply by 4, 4 
5627 + 86a — 382 
5627+ 142 — 21 
Divide by 11, YY) 22 eee 
2%— 1)824+227—3(424+38 
8a’—4a 
62—3 
fe Luertt() b= 22— 1; 62—3 





114 SCHOOL ALGEBRA. 


The following arrangement of the work will be found 
most convenient : 







827+227—8 | 6a + 5a? — 2 
827 —4a 4 
64—3 2427+ 2027— 8 32x 
62 —3 242°+ 627— Ya 
1427+ 9xr— 8 
4 
5627 + 862 — 32 +7 


5627+ 142 —21 


| 11) 222 —11 
Q2— 1 4a +8 


146. From the foregoing examples it will be seen that, in 
the algebraic process of finding the H.C.F., the follow- 
ing steps, in the order here given, must be carefully 
observed : 

I. Simple factors of the given expressions are to be re- 
moved from them, and the H.C. F. of these is to be reserved 
as a factor of the H.C. F. sought. 

II. The resulting compound expressions are to be ar- 
ranged according to the descending powers of a common 
letter; and that expression which is of the lower degree is 
to be taken for the divisor; or, if both are of the same 
degree, that whose first term has the smaller coefficient. 

III. Each division is to be continued until the remainder 
is of lower degree than the divisor. 

IV. If the final remainder of any division is found to 
contain a factor that is not a common factor of the given 
expressions, this factor 2s to be removed; and the resulting 
expression is to be used as the next divisor. 

V. A dividend whose first term is not exactly divisible 
by the first term of the divisor, is to be multypled by such 
a number as will make it thus divisible. 


Sp ple LE gs EES 1 he, Por 


wo wo DO WD S-& S&S BSB SB BS SF SF SF SF BS 
Cr es Re Ct CO OO - ered Ole Ol ee tie SC OOe se oy 6 ft 


COMMON FACTORS AND MULTIPLES. 115 


Exercise 39. 


Find by division the H.C. F. of 


427+ 382-10, 42°+ 72? — 82-15. 

22°— 627+ 524—2, 82° — 2377+ 172 —6. 
2023+ 227—182+ 48, 20a*—172?+ 482 — 83. 
Ag® — 22?—16a—91, 122° — 28 2? — 874 — 42. 
120° +42°+172—8, 242°— 522e?+ 14¢—1. 
20° + 52? —I9r74+38, 382° 4+ 22?—17274+ 12. 
8a* — 62° — 2? +1527 —25, 42° + Tx’? — 32 — 15. 
4¢°>—42?—52+38, 102?—192+ 6. 
62t—1382°+ 3277+ 22, 62*-—102'+ 42°? —- 624-4. 
22° — 82°? + 20? —2Qa—3, 4a*+32°+42—8. 


- da°—a?— 2e?4+2xe—8, 62°+182?4+ 38x-+ 20. 

. o@ + 22+ eg, 8at +22 —3827+2r-—1. 

. 6a2°—9attlla*+6 2’—102, 42°+10a'+10 2°44 24602. 
. 2e°—I11a’—9, 42°4+11l2*+81. 

ee + 102 — 1924-9 at +2274 9, 

. 2e°—382?—16274 24, 4a°4+ 22t— 282° — 162° — 322. 
. &—e—l4e4+2+1, Rigid eta a oetigs | on 
. 62°—14az?+ 6a'x—4a', 2*—az*®— aa’ — aa — 2 a. 

se20 —2a' — 3a’ — 2a, 3a‘ —a®'— 2a'— 16a. 

. 22° + Tar? +472 — 8a’, 42°4 9ar’— 2e'r — a’. 

. 22?—9az’+9aa—Ta’, 42° — 20a2? +20 ae —16.a°, 
. 2a44+ 9e'8 + 142+38, 824+ 142°4+ 927+ 2. 


92° — Ta? + 8274+ Qa —4, 6at—T2*°—102?+52742. 


116 SCHOOL ALGEBRA. 


147. The H.C. F. of three expressions may be obtained 
by resolving them into their prime factors; or by finding 
the H.C. F. of two of them, and then of that and the third 
expression. 

For, if A, B, and Care three expressions, 

and D the highest common factor of A and B, 

and £ the highest common factor of D and C, 
Then PD contains every factor common to A and B, 

and # contains every factor common to D and C. 
.. # contains every factor common to A, B, and C. 


Exercise 40. 
Find the H.C. F. of 
’+8a+2, 2+427+3, 2’ +62+5. 
x’—9x2—10, 2 — Tx — 30, 2 —11lxe#+10.. 
2—l1, e—227+1, #P-—22+1. 
62+a—2, 22°4+7Tx#—4, 2e?—Txr+38. 
vt 2ab+0?, 7@—B, a+2a7b + 2ab ? + Bb. 
xv —dax+ 4a, 2 —B8ar+2d, 32?—10ar+ 7a’. 
eta—, 2—2e—27+2, 2+327—6r—8. 
e+ T+ 52-1, 2+3e2—82'—1, 32°+ 527+2—-1. 
v—62"+1la—s6, v°—8274-192—-12, 2°—92°+ 262—24. 


O~ Or st OF Cte oe 08 NS 


148. Common Multiples. A common multiple of two or 
more numbers is a number which is exactly divisible by 
each of the numbers. 

A common multiple of two or more expressions is an 
expression which is exactly divisible by each of the ex- 
pressions. Thus, 48 is a common multiple of 4, 6, and 8; 
48 (a? —y’) is a common multiple of 3(a—y) and 8(#+ y). 


COMMON FACTORS AND MULTIPLES. 117 


149, The lowest common multiple of two or more numbers 
is the least number that is exactly divisible by each of the 
given numbers. 

The lowest common multiple of two or more expressions 
is the expression of lowest degree that is exactly divisible 
by each of the given expressions. Thus, 24(2*?—y’) is the 
lowest common multiple of 3(z—y) and 8(#+ y). 

We use L.C.M. to stand for “lowest common multiple.” 

To find the L.C.M. of two or more algebraic expressions : 


CasE I. 


150. When the factors of the expressions can be found by 
inspection, 
(1) Find the L.C.M. of 42.°0? and 60a70*. 
Siu) KT a Kb: 
60a7b*=2X2X3XK5xXa?x Ot. 


The L.C.M. must evidently contain each factor the greatest num- 
ber of times that it occurs in either expression. 


Rea A x OC XK IKK 0 6 OF, 
= 420 a®d*. 
(2) Find the L.C.M. of 
4a? + 44—12, 627-482-490, 427-102 —6. 

Agt 4¢—12=4(2?+a2—12) =2xX2(¢—3)(2#+4); 

62°—48 2+ 90=6(2?— 82415) =2 x 3(«— 8) (x%—5); 

4z’?—10x— 6=2(22°—52—3) = 2(4—8)(24+1]). 
~ L.0.M.=2x 2x 3 x (#—8)(#+4)(e—5)(22+1). 

Hence, to find the L.C. M. of two or more expressions: 

Resolve each expression into its simplest factors. 

Find the product of all the different factors, taking each 
factor the greatest number of times rt occurs mn any of the 
given expressions. 


118 


SCHOOL ALGEBRA, 


Exercise 41. 


Find the L.C.M. of 


a I 


EF FF EF >» SF S we we 
mia) OU He OO, (Or et 


ce  .00- wets Co) OU hare OO SN 


24, 32, and 60. 

24 a7x*, 60 a0*2?, and 32472’, 

xv’? —2Qeyty’ and 2 —y’. 
v’—4e+4, a?+4a2+4, and 2 —4. 
sta’ and 2? — a’. 

Ytarta’, v’—a’, and 2’— a’. 

x" (%2— 3) and 27— 52-46. 

e+ Ta+12 and 2 — 92’. 
v—Ta+10, 2—42—5, 2? —ax—2. 


1—32—421—42—527, 1292-1 


. 62+ Try —3y’, 32?4+ llay—4y’, 22°+ llezy+12y’. 
. 8—14a+6a’, 4a+4a?—38a', 40°+2a?— 6a". 

. 62+72°—8a, 82°+142—5, 62°+392+45. 

- 6ax+9bx—2ay—3by, 62° + 3a%—2xy — ay. 

. l2Zax—9ay—82y+ 67’, bax+ 3ay—4ay—2y’. 
. 272?—a’, 62? +ar—a’, 152°—5ax+8ba—ab. 


. @—)l, 2e?—ax—-1, 32?—2—-2. 


Case IT. 


151, When the factors of the expressions cannot be found by 
inspection. : 

In this case the factors of the given expressions may be 
found by finding their H.C.F. and dividing each expres- 
sion by this H.C. F. 


COMMON FACTORS AND MULTIPLES. 119 


Find the L.C.M. of 


62° — llay + 27° and 92° — 22 xy? — 8 y’. 


62°—11] ay +2 7° 9a°—222y'— 8y’* 3 
62°— 82°y—4 27’ 2 

— d7y+4 277127 18 2°—44 x7'—16 7? | 

— 8ay4+4a77427' 18 2°—338 a’y+ 6y’' 


lly) 33 v’y—44 zy’—22 y° 
82° — 4ay— 2y*?|Qr—y 


.. the H.C. F. = 32? —42y —2y’. 
Hence, 62°—11lw’y+2y?=(22—y) (82°?—4xy—2y’), 


and 


92° —22 ay’ —8y=(82+4y) (82°—42y—2y’). 


.. the L.C.M. = (22—y) (84+ 4y) (82 —42y—2y’). 


Exercise 42. 


Find the L.C. M. of 


dh 


CO IF TP wD 


- = & 
wo wm e OS 


62° — Tax’? — 20a7x, 32? + ax — 4a’. 

32° — 182? + 234 —21, 6a? + 2? — 4424 21. 
382° — 8ay+ay—y’, 4e°— ay — 827’. 

e& —2c-+c, 2c — 2c —2c— 2. 
v—82r+3, 2°—32°+ 212-8. 

a —6a74+12 az? — 82’, 2a? — 8axr+ 82". 
22+27—12272+9, 22° —72?+122-9. 
te — 227 — 5, (2? + 1227+ 102+ 5. 

zt — 1327+ 36, 2t— 2’ —Te’+ +6. 
22°1327—Ta—10, 4a°—42?—9x4-+5. 


. 1222— 2? —802—16, 62° — 22? — 13824 —6. 
. 684+ 2? —5r—2, 6a® +52°—382—2. 
ee Or + 267 — 24, a — 1907 4-47 2 — 60, 


CHAPTER IX. 


FRACTIONS. 


162. An algebraic fraction is the indicated quotient of 


two expressions, written in the form Z 


The dividend a is called the numerator, and the divisor 6 
is called the denominator. 

The numerator and denominator are called the terms of 
the fraction. 


153, The introduction of the same factor into the divi- 
dend and divisor does not alter the value of the quotient, 
and the rejection of the same factor from the dividend and 
divisor does not alter the value of the quotient. Thus 


ce ap SLE 3 Lorene . It follows, therefore, that 


4 2x4 7 422 

The value of a fraction is not altered uf the numerator and 
denominator are both multiplied, or both divided, by the 
same factor. 





REDUCTION OF FRACTIONS. 


164. To reduce a fraction is to change its form without 
altering its value. 


Case I. 


155, To reduce a fraction to its lowest terms. 


A fraction is in its lowest terms when the numerator and 
denominator have no common factor. We haye, therefore, 
the following rule ; 


FRACTIONS. 121 


Resolve the numerator and denonunator into ther prime 
factors, and cancel all the common factors; or, divide the 
numerator and denominator by ther highest common factor. 


Reduce the following fractions to their lowest terms : 


(1) 38.a'bict _ 2 K 19a*b®c* _ 26?c” 


57abe =—8x19a°be? — Ba 


(2) fee (a — a) (ai t-ar+a") a +ar+ x 
ve—x (a—xz)(a+2) ataz 

(3) a+fa+10_(@+5)(a+2)_a+5 
at+d5a+6 (a+3)(a+2) a+3 


(4) ee oe 20 8 (22-38) (82-2) sa-+2 
827—2x%—15 (2%—38)(47+5) 424+5 
e—4e7+42—1 

eee 

(5) v—2e+424—8 

We find by the method of division the H.C.F. of the 

numerator and denominator to be x — 1. 

The numerator divided by z—1 gives 2 —3x+1. 

The denominator divided by «—1 gives x’ —2-+8. 

See 42-1 abet) 

See Ae 8 at — e+ 3 











Exercise 43. 


Reduce to lowest terms : 











i, & ab> 4. 42mid. 7 34 any? 
" Jab " 49 mn? 5lavay' 
2 306°C P 30 xyz 35. a°b'c? 
" 15a°b*e? "18 2%2? ~ Baibic 
3 26 ay* _ 21 mint «BB abict 

Bony 28 mp 87 a'b*¢ 








a bap 


12. 


13. 


14. 


15. 


16. 


1? fe 


18. 


28. 


29. 


30. 


31. 


SCHOOL ALGEBRA. 


ZO eae 
4 (a? + ae + c’) 
2+ 22y+y' 

aa eT ioe a oh 
xv’+ 2a—15 

piroeeh St zers wi i.e) 
227— 132+21 


a —~ 27 —20 5 
92°— Tx—16 


ae? —62—4 


32° — 8a2+8 


Geena wed 
e+t427—5 


8 ee ad 
82° —427—a21+2 


zt — 182’? + 36 


19. 


20. 


21. 


22. 


23. 


24. 


25. 


26. 


27. 


32. 


33. 


34. 


427+ 12axr+ Ia 
82° + 27a’ 

v—y —- 2yz— Zz 

w+ 2ry+y—2 

at 4. aa? 
yp 

20+ 17a+21 

3807+ 26a-+ 385 

(a- bye 

(a+tb+ep 

creel I 

Y—x 

(oa) — oe 

(7+ by—@ 

(a+6) — (e+ da) 

(a-+c)?— (6+ ad) 


(a+ec)—6 
4 ag? — (a? oe 


Reduce by finding the H.C.F. of the terms : 


38° + 17a? + 222448 
62° +25 27+ 2582+ 6 
2° — 3a — 1b ea Zo 
et T2’+5a—25 
22° 2) aie 
82°+8e2+2—2 


a + Aa) — Bae 
z'—a2+ 82-8 


FRACTIONS. ees 


Case II. 


156. To reduce a fraction to an integral or mixed expression. 





(1) Reduce = 


; to an integral expression. 














Pata tetl (§ 108) 
(2) Reduce aa to a mixed expression. 
2—1 |“z+1 
+e 2t—etl 
—x—x 
eo) 
x+1 
—2 
x —l1 ; 2 
y Sek ei ee 


Note. By the Law of Signs for division, 


2 
a 


ee, 
2 





2 
——— and = ; 
z+1 —(# +1) x+l 





The last form is the form usually written. 


157, If the degree of the numerator of a fraction equals 
or exceeds that of the denominator, the fraction may be 
changed to a mixed or integral expression by the following 
rule: 


Divide the numerator by the denominator. 


Note. If there is a remainder, this remainder must be written as 
the numerator of a fraction of which the divisor is the denominator, 
and this fraction with its proper sign must be annexed to the integral 
part of the quotient. 


124 SCHOOL ALGEBRA. 


Exercise 44. 
Reduce to integral or mixed expressions : 


4a°+ lat 38 3 29 eee 

















3 SE EEERASREEP SES ane 
Ag atod. 

9 ae ete en eA 9 Sat Adee le 

; 32 a+t4 

fy, wets. 10, © eee 
ety a+2z 

ie es he iL ae 54 
t—Y yp 

P aes Hh 12 3 iy pa ae 

as gt ee 

” Shae Ee 13, 4@+6ar+9at 
ety 22—3a 

7 v+81 14 one 4 ate 

Wine in eg @ ae 

Case III. 


158. To reduce a mixed expression to a fraction. 
The process is precisely the same as in Arithmetic. Hence, 


Multiply the integral expression by the. denominator, to 
the product add the numerator, and under the result write 
the denominator. 








(1) Reduce to a fraction “— i +26. 
gr 
G38 a ee oe 
z—4 x—4 
ae ore ee 


x—4 
De ee Ly 
x—4 


FRACTIONS. 


(2) Reduce to a fraction a— b — 


125 


a Te nO enue 


até 


_(a@—6)(a+6) — (a — ab — 0) 


atbéb 


_@—B—a+ab +P 


ee ab . 
a+b 





a+b 


Norr. The dividing line between the terms of a fraction has the 


force of a vinculum affecting the numerator. 


If, therefore, a minus 


sign precedes the dividing line, as in Example (2), and this line is 
removed, the numerator of the given fraction must be enclosed in a 
parenthesis preceded by the minus sign, or the sign of every term of 


the numerator must be changed. 


Exercise 45. 


Reduce to a fraction : 

















1: ee 8. 
9; ee 9 
22 
2ab 
Le b— 10. 
Ps atb 
2 
4. Aes 11. 
. x—2 
pe | el 12. 
a+6 
| gpa Ta 
a 
T atx iss a 14. 


. @-—art+2— 





ial 
CN eSios 
co x—3 


3 


ata 





Vtarzt2— 





BIN at Es 


es eR By 


126 SCHOOL ALGEBRA. 


CasE IV. 


159. To reduce fractions to their lowest common denominator. 


Since the value of a fraction is not altered by multiply- 
ing its numerator and denominator by the same factor 
(§ 153), any number of fractions can be reduced to equiva- 
lent fractions having the same denominator. 

The process is the same as in Arithmetic. Hence we 
have the following rule: 


Find the lowest common multiple of the denominators ; 
this will be the required denominator. Divide this denomi- 
nator by the denominator of each fraction. 

Multiply the first numerator by the first quotient, the sec- 
ond numerator by the second quotient, and so on. 

The products will be the respective numerators of the 
equwalent fractions. 


Norse. Every fraction should be in its lowest terms before the 
common denominator is found. 


ox 2 5 : 
(1) Reduce hae a and Ba to equivalent fractions 


having the lowest common denominator. 


The L.C.M. of 4a?, 3a, and 6a? = 12a’. 
The respective quotients are 3a, 4a”, and 2. 
The products are 9aa, 8a?y, and 10. 
Hence, the required fractions are 


Jax 8 ary nid Reo 
12a? 1243 12a° 





in 2 3 
2) Red —______, ——_—___., —_______ t 
fee ccs e+ 5e+t6 +4748 3 eo i 
equivalent fractions having the lowest common denom- 
inator. , 


ibe 


FRACTIONS. ey 


ee eee Se 
e+5e+6 @442743 2+22+1 
in 1 2 3 
~ (w@+3)(e +2) (@ +3)(@41) (@+1)(@+1) 
. the lowest common denominator (L.C. D.) is 
(w + 3) (@ + 2)(@ +1) (a + 1). 
The respective quotients are 
(w+ 1)(@ +1), (w+ 2)(@ +1), and (# + 3) (a + 2). 
The respective products are 
1(@+1)(@+1), 2(@+2)(e+1), and 3(@+3)(a + 2). 
Hence the required fractions are 
(w +1)(« +1) 2(w + 2)(e+1) 
(w + 3)(w@ + 2)(a +1)? (w@+3)(@ + 2)(~@4+ 1)? 


3 (x + 8) (x + 2) 


(x + 3)(x + 2) (a + 1)? 


Exercise 46. 


Express with lowest common denominator : 









































ae ak — 2 Oat. 4 Ft. 7G 
3a Jax MAgew? Sa — ¢ 
1 Bo 5 vty’ Layiphi 
+2 4+8 20%7—4y? Sa+2y 
a a" : 6 +2 sid 
t—a 2a Ve ey ee 

7 1 1 2 

ety a—-y er y¥ 

8 1 1 3 

T1422 1—42 1-22 

9 5 ’ 7 ; 3 

yt) Reta Gaeta 


1 
w—9r+18 2?—102+ 24 


128 SCHOOL ALGEBRA. 


ADDITION AND SUBTRACTION OF FRACTIONS. 


160. The algebraic sum of two or more fractions which 
have the same denominator, is a fraction whose numerator 
is the algebraic sum of the numerators of the given frac- 
tions, and whose denominator is the common denominator 
of the given fractions. This follows from the distributive 
law of division. 

If the fractions to be added have not the same denomi- 
nator, they must first be reduced to equivalent fractions 
having the same denominator. (§ 159.) 

Hence, to add fractions, we have the following rule: 


Reduce the fractions to equivalent fractions having the 
same denominator; and write the sum of the numerators 
of these fractions over the common denominator. 


161. When the denominators are simple expressions. 


Sun pe BOAO. Fa— oO eee =a 
1) Simplify ——— — ———— +. ——_. 
The L. C. D. = 12. 
The multipliers, that is, the quotients obtained by dividing 12 by 
4, 3, and 12, are 3, 4, and 1. 
Hence the sum of the fractions equals 


9a—12b_ 8a~4b+44¢,a—4e 
12 12 12 

_ 9a—12b6—(8a—4b + 4c) +a—4e 
Ye 

_ 9a—12b-—8a+4b—4ce+4a—4e 
12 . 

_ 2a—8b—8¢ 

12 
_~7—4b—4e 


6 


FRACTIONS. 


The oaks work may be arranged as follows : 


The L. C. D. = 12. 
The multipliers are 3, 4, and 1, respectively. 


3(3a—4b) = 9a—12b = lst numerator. 
—4(2a—b+c)=—8a+ 4b—4c= 2d numerator, 
l(a — 4c) = a —4c¢c= 3d numerator. 

* 2a— 8b—8c 


129 


or 2(a—4b—4c) =the sum of the numerators. 


.“. sum of fractions = 





2(a—4b—4c) a—46—4e 
1 6 


Exercise 47. 


Simplify : 














3 6 eS Poh aay 
tz@—5 38242 ,4¢+1 5¢—10 
aes 3 4 12 
22+38 
9 6 12 3 


24+3,2+38 182%4+5 w2«-8 


2x 4x 82? x 








oe 29-1 oes Die tates 








2 5) ron eer 12 


2 2 ae ey ee 
A Neral ek Oe ee whi 4 at 
b ab a ab 











4 16. h 12 3 


pega a al OE 7. 
e 2 7 5 dia 8 














fe eee Lelie) we PE ees 





1380 SCHOOL ALGEBRA. 





Qr—6 8a—4 , 56xr— 48 
1 Osa soe Bi 
5a 15x as 45a ; 
11 Liay Nt Syhaao! L6at— B 
: ay? ay? ay 











lone eluate ot 8 ee 
2a'y  Oy'2. Daz Ave eae 


162. When the denominators have compound expressions. 


a 


—b atb @—B 
The L.C. D. is (a — b)(a + 8). ; 
The multipliers are a + b, a—}b, and 1, respectively. 


(a+ 6)(2a+b)= 2a*%+3ab + 0? = Ist numerator. 
—(a—b)(2a—b) = — 2a? + 3ab — 0? = 2d numerator. 
—1(6ab) = —6ab = 3d numerator. 


0 = sum of numerators. 











*, sum of fractions = 0. 


: : Pee ie ees 
2) Simplify ~ 
(2) bra ge Oa oi 4 


The L.C.D. is (w — 2) (w — 3) (a — 4), 
(«—1)(@—3)(@a—4)= a — 8a%+19x%—12 = Ist numerator. 


(w — 2)(w@— 2)(a@—4)= a — 82a? + 20x—16 = 2d numerator. 
(x — 2)(e@—3)(e—3)= a — 8a? + 21% —18 = 3d numerator. 











32° — 242? + 60a — 46 = sum of numerators. 


32° — 244? + 60a — 46 


~ gum of:aractions = ——— ee 
(@ — 2)(# — 8)(«— 4) 


Exercise 48. 














Simplify : 
1 1 il 2, 
1. ; eo nes : 
Rey Oo eB lta 1-2“ 
2, l el 4 ty ee 


Itz l—z x—y (x£—y) 
































Hint. Reduce the first fraction to lowest terms. 

















FRACTIONS. 131 
pee ty)" gy Se Sy tay 
ary («@t+y) a—y “ty w—y’ 
1 1 x x Ve 
ee ee eS 10, : 
PEGA 2a(a =) eee, 1a, ve+e2 
7 od tes Se een ae ling 11 it, Cage Bie ; 
; aa l—#4+27 Sa Sgt os Oe bane te ap 
Rela oc) 12 Doe awe ey 
mere a — 27¢ OZR Sih a 28S a 
ise fea ace, 
e+2y xw-—2y 2—4y7* 
i a 
Veet ay Le ae — 
x ik 1 
15. 1 = 
x—1l aes 
Fee aume Mmaber wn nirabin 
~atb (a+by¥ (a+6) 
v7, Sy 22 # ey) 
Tee ged kar YF) 
Hint. Reduce the last fraction to lowest terms. 
18. 3 A Gee eds sDa* 
z—a (#a—a) (x—a) 
19. alae A cl a 
x—2a 2 —8a' 2#+2ar+4e 
?_94¢+8 x— 2 T 
20.0. We aes eth Sy Ee 
etl b AlsGand x+1 
8 x—l ve+tae—s 
TNE BOS Ope Sok raat oe a We Eh Pe AR 
Oe oe Be Perea xv — 27 
29 Te Oe 10) ke 1 
ee Geer te LO biont te 


132 SCHOOL ALGEBRA. 


wi—Ssan+6a  r—Ta 
x—S8ar+15e x«—Dd5a 
iy 3 See oe 
xa—-2 #—82r4+2 w#—4244+3 


23. 





24. 





Hint. Express the denominators of the last two fractions in 
' prime factors. 


3 I 
w@-Tatl2 d@—4a+38 av—5a+4 

b Gini se SH ae ee 
l0¢?+a—38 2¢+7Ta—4 

Bibs 3 u 


Q—x2—-62 1l—x—22* 


163. Since 2 a, and pan a, it follows that 


The value of a fraction is not altered of the signs of the 
numerator and denominator are both changed. 

It follows, also, by the Law of Signs, that 

The value of a fraction is not altered if the signs of any 
even number of factors in the numerator and denominator 
of a fraction are changed. 


164, Since changing the sign before a fraction is equiva- 
lent to changing the sign before the numerator or the 
denominator, it follows that 


The sign before the denonunator may be changed, provided 
the sign before the fraction is changed. 


Note. Ii the denominator is a compound expression, the beginner 
must remember that the sign of the denominator is changed by 
changing the sign of every term of the denominator. Thus, 


x x 


a—2 x~—a 





These principles enable us to change the signs of frac- 
tions, if necessary, so that their denominators shall be 
arranged in the same order. 


FRACTIONS. 133 


3 22—3 
soe 
2%—1 1—4z 
Change the signs before the terms of the denominator of the third 
fraction, and the sign before the fraction we have 





(1) Simplify 2— 
xv 


The L.C. D. = 2(2a% —1)(2a2 + 1). 
2(2e@—1)(2%+1)= 827—2 = 1st numerator. 
—3a(2%+1) =— 6a?—3e= 2d numerator. 
—«(2%—3) = — 207 +3x2=3d numerator. 
—2 =sum of numerators. 
Sean of the fractions =— —_» “> 
x(2%—1)(22 + 1) 
(2) Simplify 
1 i 1 
a(a—b)(a—c) 6b(6—a)(b—c) c(e—a)(e—)) 
Nort. Change the sign of the factor (6 —a) in the denominator 
of the second fraction, and change the sign before the fraction. 
Change the signs of the two factors (e—a) and (¢ — 6) in the de- 
nominator of the third fraction. We now have 


soe ae oe MS 
a(a—b)(a—c) b(a—b)(b—c) e(a—e)(b—e) 
The L.C.D. = abe (a — b) (a — c) (6 — ¢). 


be (b —c) = be — be? = lst numerator. 
~—ac(a—c)=—a@e+ae? = 2d numerator. 
ab(a—b) = a*b — ab? = 3d numerator, 


ab — ae — ab? + ac? + be — be? = sum of numerators. 
= a? (b — c) — a(b? — c*) + be (b— 0), 

= (a? —a(b +c) + be] [b — ¢], 

= (a? — ab — ac + be}[b —c], 

= ([(a? — ac) — (ab — bc) |[b — ¢], 

=[a(a —c) — b(a—c)][b —c], 

= (a — b)(a—c)(b— Cc). 


(a—b)(a—c)(b—c) _ 


* gum of the fractions = Eh. 
abe(a — b)(a—c)(b—c) abe 


134 


SCHOOL ALGEBRA. 


Exercise 49. 


Simplify : 


1. 


10. 


1 Rs 


12. 


13. 


2 
x x x 
-- 


v?—1 ah. lee 








— va 

ae swuien o LD 
2a SISOS ALO atone a 
a—b 2a a+ ab 


b APRN LT TOT ST 





























3 5 24 —7 
2 1-22 481 
psi ty 2 1 bye bias 
(ae Gives v—a@ 
is pen ee ey sabe, 
r—y Y+ayty y—x 
1 2 1 


(@—2)(e—8)' @—l)@—a)' @-N@—2) 
be ac ab 


CENCE CECE 


b+te ate a+b 
Seles 
pee. a eg nes Se re ee es et a ees 8 
(a —b)(6—c) te Res ea 
1 1 Ne da 
c@—y@—-)  ¥y—NY—D rye 
a? — be 67 + ae 4 ce? + ab 


(Gaby (aan (6—a)(b+c)' (e—a)(e+b) 


FRACTIONS. 135 


MULTIPLICATION AND DIvIsIon oF FRACTIONS. 
165. Multiplication of Fractions. 
ae 
Ee ons 
i b da | 
quotient 7 by a, and divide the result by 0. 


The expression means that we are to multiply the 


From the nature of division (§ 76) if we multiply the 
dividend ¢ by a, we multiply the quotient 4 by a, and 
obtain oe if we multiply the divisor d by 0, we divide 


the quotient ee by 6, and obtain a Hence, 
LIBS A ceageet 
bd bd 


Therefore, to find the product of two fractions, we have 
the following rule: 


Find the product of the numerators for the required 
numerator, and the product of the denominators for the 
required denominator. 


166. In like manner, 


and so on for any number of fractions. 


2 2 
Again, eae 
In like manner, 
Xe OF 
(5) “FF 
167. Division of Fractions. If the product of two numbers 


is equal to 1, each of the numbers is called the reciprocal of 
the other. 


136 SCHOOL ALGEBRA. 


The reciprocal of 5 is a 


a 
b ameba 

f ON 
ee eee ab 


The reciprocal of a fraction, therefore, is the fraction 
inverted. 


Qi deere | 
ince a he 
od BO va tol oun 
eid 


To divide by a fraction ws the same as to multiply by rts 
reciprocal. 
- To divide by a fraction, therefore, 

Invert the disor and multiply. 


Nore. Every mixed expression should first be reduced to a frac- 
tion, and every integral expression should be written as a fraction 
having 1 for the denominator. If a factor is common to a numera- 
tor and a denominator, it should be cancelled, as the cancelling of a 
common factor before the multiplication is evidently equivalent to 
cancelling it after the multiplication. 


2a%b \, bed 5.abrc 
Scd* 5ab “Sec 
20% bed. 5ab’e 2x6~x 5a°bied ine 
3cd?°5ab ~~ 8a2etd? 3x5x8ar%bedt 223 





(1) Find the product of 











(2) Find the product of 
re AE is ay" ey 
x’ —d32y+27 eee * (oe 
oy" ye Y= 2y" — «ey 
—3ay+2y? x4 ay ae y)? 
(7—y)(@+y) x Ya = 2 Dy a(e—y) 
“(e—y)@— 29) a@+y) ~@-Ne-H) 
Swine 
Cae 





FRACTIONS. 137 


Notz. The common factors cancelled are (x — y), (a + y), (w—2y), 
w, and w— y. 








(3) Find the quotient of aaa - 
Pie ~ OB 5. an ye (4= 2) (a + 2) 
(a—2)? a?—2? (a—2x)(a—z) ab 
EY x(a + x) 
b(a— 2) 


The common factors cancelled are a and a— wz. 


Exercise 50. 











Simplify : 
1, Lax, aay Ti Sar x 4° 
fo Spl NS iad e—B dv+ab+e 
» 2 arb?c® , 20 mn? 11, DEY Be H+ 8Y 
4amn  2la’te ey x+y 
3 6 abe? Sm?a* 12 san Gi res 
 Tmay” 8c! al(a+b) al(ai—2) 
4 San 42° 13 Ga ke 0 oe 








8a%y? ~ 15 mn f+ 4ax~ az+4a* 
16 ait? . 40° wu, Say er ey. 























Bl ary? ~ Baty @—y~ wy 
&: Tay Be 3522 | 16. z+ a’ a+3a 
1277 - 3642’ Cm OU Beto 
7 aa ge 16 AE ea hc eae Meas ia 
ety «x+y OLB a&—ab+ Fb 
3 32° — 2 2a 17 xz? — 1 re ae 
a 227 —4a " Ay —-5 ®t 2x8 








(ee) 


8 2 82 —6 S a 
9. - ; 18. (1— ES, peer 
5a — 10 * 4x ( a xy 


30. 


31. 


SCHOOL ALGEBRA. > 


eae 
y au—y 

ae, aR ATH aay 
y 42° + 22ry+y’ 


acerca 


ary xv’ +? 


8a’b .. od 4ab.. bed — cd? 


GSB bed «A (paabay 


u@=), @=¥)_, = 





aa+y) #+ay+y («+y) 


a 
a 


a? + ab —ac ab —b? — be 


(EO) Ce ene c—a—b 


a—(b—cy ce—(a+by ac — a? + ab 


(2 —a)'— 6 at — (6 —a)' | ax + a! — ab 
(c—b?v—a@ 2#—(a—byY be—ab+O 


OOO 0 Ce he 


“v+2ab+0?—c a—b+e 


ea ee 
x 


a(%+1) xu? +- 4x 
_2ax’ + Zatz v—a v, z+ a 
(c—ay(e@+a)y 2(2+0) ax 
a? — 6b? — c?— 2be ae 2c ) 
a’ — ab—ae at+bte 
a+ab?t+ bo até 


2573 
x ore x. 
a’ — b§ a+b a 











30. e+ Tay + lay, n+ cy — 2y7 | 


Y+5ayt+6y #+32y—-47 


FRACTIONS. 139 


168. Complex Fractions. A complex fraction is one that 
has a fraction in the numerator, or in the denominator, or 


in both. 

















eee 2 
(1) Simplify ey 
8a 38x e242 7-1 32 4 
foe 4c 1-1 AEE Teele Aig a] 
4 
So Ae 
ar ee 


Nors. Generally, the shortest way to simplify a complex fraction 
is to multiply both terms of the fraction by the L.C.D. of the frac- 
tions contained in the numerator and denominator. Thus, in (1), if 
12a 


we multiply both terms by 4, we have at once EE 
H bid 





Qe be 
(2) Simplify 
a—-#L at+az 
The L.C. D. of the fractions in the numerator and denom- 
inator is 
(a—x)(a+2). 
Multiply by (a—2)(a+2), and the result is 
(a+ 2)?—(a—2) 
(a+a)+(a—zy 
_ (a + 2a + 2”) — (a — 2axr+ 2) 
(a? + 2ax + 2”) + (a — 2axr-+ 2’) 
me -- 20% + me —@V+2axr— 2 
VC+2%arte2t+e¢—2ar+2’ 
etter 
2a? + 22° 
_ 200 
C+ 2? 


140 2 SCHOOL ALGEBRA. 


(3) Simplify 


Fane 
z s 
he £ ie CAG Eee 
lpa2+ (l+2)(l—2+2’)+2 


x 
ee ert eo 
lt+a2+2 
as a(1l+2+2°) 
1+2+2°—(*x#—2+ 2°) 
te pt 
1+ 2 


Nore. In a fraction of this kind, called a continued fraction, we 
begin at the bottom, and reduce step by step. Thus, in the last 


example, we take out the fraction me 


xv 


1+2+———— 
1l—a#+2? 


, and multiply 


the numerator and denominator by 1 — 2 + 2?, getting for the result, 


2 2 3 
Saat Nocera which simplified is oer ee 
(l+a)\(1l—v2+4+2%)+¢e l+a+2° 


Putting this fraction in the given complex fraction for 


v 


ye eee ee 
l—a+2? 


x 
1 #o wv +e 
l+toat+a2 


we have 


Multiplying both terms by 1+ 2 + 23, we get 
a(1 + a” + 2%) 
l+e2+2—a2+a%-—23 
ss Ee 
1+ 2 


Simplify : 











FRACTIONS. 


Exercise 51. 




















10. 


Ti 


12. 


13. 


14. 


2m+n_ 4 
mtn 
ae 
m+n 
vty’ 
oy 
et —ay ty 
dito Y 








1 a 


i 





GD) 





Se ee 
Rae yp ITE 


1 1 1 
ere 


ab sae 


ab 


— 





141 


142 SCHOOL ALGEBRA. 











15. sae 16. —__#+4 
SL rT. tT Ye 
1 a 
ve = re 
get 
17 ee ee ee 
ee ar y (xyz + x -+ 2) 
Tie 
Z 
gat tet 
18. 83 .abe ek Ce 2 
bc+ac+ab a te 
8 ae 





19. Gee) es oe 


Mn 6 0 90 


20. ese See 
arty #+y7* LL — ae 








1 1 
=a 
pews, G+ e—a’ 
e 1 RET STRESS i> 
Re Lage Lae 1 (1+ 2be ») 
a b+e 


Exercise 52. 
EXAMPLES FOR REVIEW. 
1. Find the value of Va? + B® +  — (a—6b—e)’, when 
a=? bS=—2, and ¢c=4 
3° + LO mide J ee 
82°+ 1382°+17%+6 





2. Reduce to lowest terms 


Simplify : 


1 3 co 4. 2-38 
* @—3)@—-2) @-N@—-s) @—-Ne—2 


Ua 


FRACTIONS. 143 


; ee eye Yi + re ad A ete ws Ba ; 
eee 2) Y—2)\(y—2)) eae Hy 


z l—2\. x , 12) 
Geet a JE Gace 2 ) 
Af oan 1 ay 3 
e\a—-e 2x£4+2¢ x +ex—2¢e 


xt — y* x? 
° ae a thc Se aa | 
vty? Car 3 =) 


Gere a -2) 8 (a'—z—2) 8a 


xv—x—2 v’ta—2 x’? —4 


 (et2y “+s » 
(oer +5) Ga) 


Meaaite3)(4 = eas 





























toa ty Pa (y—2y 

wt xy + y’ ae (e+y) 

26 0—C As 26—c—a of Pes xOd 
(@—B(a—)' @—e(b—a)' C—a)(e—B) 
s! 2 Leah eels 
ttl («+2)(2+1) 7 aes ead ae 

Sra Oe cea ay is a 
+2 (¢#+1)(#+2) reel DE Ay andl 











pee oe pl 
ee ed ty Oe a") 
P—8y y vm ayty xe —y). 
: a(a—y),.2+22ay+ 47° e+y?* 
eee co eta} at be 


Me ads 0 be ac ab 


CHAPTER X. 


FRACTIONAL EQUATIONS. 


169. To reduce Equations containing Fractions. 





(1) Solve eR eS 9 


5) 11 
Multiply by 33, the L.C.M. of the denominators. 
Then, lle —32 +3 = 332 — 297, 
lla —3a— 332 = — 297 — 3, 
— 252 = — 300. 
“. & = 12. 


Nore. Since the minus sign precedes the second fraction, in remov- 
ing the denominator, the + (understood) before a, the first term of the 
numerator, is changed to —, and the — before 1, the second term of 
the numerator, is changed to +. 


Therefore, to clear an equation of fractions, 
Multiply each term by the L.C.M. of the denominators. 
If a fraction is preceded by a minus sign, the sign of 


every term of the numerator must be changed when the 
denominator 1s removed. 


xa—4.a4-—5b 2~T ¢—8 














¢— 5 iy 6 6 ee 

Note. The solution of this and similar problems will be much 
easier by combining the fractions on the left side and the fractions 
on the right side than by the rule given above. 


(2 — 4) (x — 6) ~(w@— 5) _ (w— 7) (e@—9)—(2@— 8? 
(a — 5) (a — 6) (a — 8)(« — 9) 


(2) Solve 


FRACTIONAL EQUATIONS. 145 
¢€ 


By simplifying the numerators, we have 


coe bee ed HS SN ete A 
(7 —5)(w—6) («—8)(a—9) 


Since the numerators are equal, the denominators are equal. 
Hence, (x — 5) (a — 6) = (w — 8) (aw — 9). 


Solving, we have 2 = 7. 


Exercise 53. 









































Solve: 
‘; Pee eel aes 5g 
4 3 5 
2. 2 in 5 Le 8 
6 4 
s Seri, ler? 8e-1_ir—l 
3 9 2g 6 
feo — 2 lle +2 272 
4 eee ee oe I Se 
a 2 14 3 
62—4 18 —42z 
: —2=> A 
5 3 3 +2 
62+7 ,Tx—18_ 2xr%4+4 
eet gee 
9 3 27 5 
7 too 00-7 36x +115 Al 
peel a 14 56 56 
a pees i Sa 5), 1 
2 a 1 4 
ae2-—-1., 2a+1 Reb 1 a— 1 
9. 11 — =10— 
Tea ce on a ae 
Tx—4 ,382%—1 5(a—1)_ 3(8x2—1),2 
ee ee ee 
9 e 5 6 20 v7 


146 


oe 


13. 


14. 


15. 


16. 


A Bs 


18. 


19. 


25. 


26. 


27. 


28. 


SCHOOL ALGEBRA. 





6¢ —2i@—1_2%42—1)_9e—5_ Me—2, 99 
4 5 4 Q 
l0z+1l 122-18 4, 7-62 | 
6 3 4 
3 5 9 1 
yi! 3-H 2y—-H 2 
Qi ge 57 8 eee 


Om 21) 8iv 1) ae 





10—"2_5#—4 

















eis a el 

6—Tx 5a 2a—1 42?—1 22+4+1 

an ped: aii 3) eter oh 21 9—2¢ .24— (toes 

8+2x2 13—22 OE ge 2} zx+1 a2’ —] 
¢ 2 

iE cael pent? 29. 622) Be 2 oi 6 

















Feces coeees c+Q x—-2 #- 


oo 


x w-—bdxr 2 4 ttl 2 
23. SSS ee 
8 8x—T 8 loch gah nee 1—2 


2ea+tl Ta—-1 22?—38x—45 


Te ee 











WO es WaG ae 42? — 4 


e7e#+l #@fetl_» 
x—]l xc+t+l 

9r+5 Oi ORL a oe 

6(2—1) 18(a?— 1) 9(#+1) 


x. 














eS “f 7 ea ees 2: 
et+2 £43 a#?+52+6 
1 1 1 1 








FRACTIONAL EQUATIONS. 147 


170. If the denominators contain both simple and com- 
pound expressions, it is best to remove the simple expressions 
first, and then each compound expression in turn. After 
each multiplication the result should be reduced to the 
simplest form. 





874+5; Ta#—8 474+6 
1) Sol eR gene 
OO) IS re aa ere aa 


Multiply both sides by 14. 
Then, Be pnee Ce) 8s ai, 
32+1 


Transpose and combine, ee ax 7; 
v + 


Divide by 7 and multiply by 32 +1, 
T7e@—3=32+1. 


c=. 





Sat be 
2) Solve es 
(2) 4 4 10 
Simplify the complex fractions by multiplying both terms of each 
fraction by 9. 





Multiply both sides by 180. 
135 — 20x = 45 — 14a + 54, 
— 62 =-— 36. 


“a2=6, 


Exercise 54. 











Solve: 
1 Ae 38 22k 2e—-1 
rh Wee 5 
9 ES Na eS 


ee Beals oer wae! 


148 SCHOOL ALGEBRA. 
lO e--l ySloa tes Eb ara 
18 132—16 9 


Oa bent © salon oie 
9 62+3 5 














ae aay 


154 2 oka ae 
62-41. Se — doe 

















6. —_ = 
15 7x—16 4: 

7 Lig pon ee 32a 

14 28 2(82+7) 

~ At Oe ees 42 le ee 
9 liz—82 "38.12 36 

9. Sica ca here it’s 2274 
15 14(¢—1) 21 6 105 


10. = 








171. Literal Equations. Literal equations are equations 
in which some or all of the given numbers are represented 
by letters; the numbers regarded as known numbers are 
usually represented by the first letters of the alphabet. 

(1) (a—2) (a+ 2) = 207+ 2axr—2’. 

Then, a? — a7 = 2a? + 2axr— 27, 


— 2ar= a’, 


Exercise 55. 
Solve: 


1. az+2b6=36bx+4a. 3. (a+2)(6+2)=-2(x—- 0). 
2. #+6'=(a—-2*. 4. (x—a)(a+b)=(x—b)(x—C). 


a 


7 


FRACTIONAL EQUATIONS. 149 





























eee eed | be. 
6b Cc 


20a — bx 


26. 
5a 


27. 





b 2¢ 





ie 


az b—« 
































Rea 2,D 17, 22% — 22 ——a\* 
ataz cetcz bbe GA Sar eh 
e+d _m—« “ a2—a_(x—by 
ab+bz an+n2x EBS it OG a 
tt+2 m+n meen 
x—-2 m—n 
10g eee Se 
MEN _m—N me b 
2+2 2-2 m 
a+br__c+dz ax—b axtb 
b d 
Re aE oT tg teen ra 
6a+a_3824—)b a 6b 
4z+6 2x—-a b a 
a a1, ———==%, 
a 
az—b_ br+e_ 7, = 
c a 
ORS 7 Sn aside. 
a— 3 b 
ee re 
mee Oe Ord a) = a : 
a—b atb @—l? 
Lr) RZ a9) Boe 8. GTO 6 
Sea Pitice bite ax—e 
i 94, tta_x—b_ 2(a+) 
a b—a b+a x—-a «+6 x 


9e—azx 6d—cr _46 


OC a 2d 
a(b—x) _ 
34 a. 


150 | SCHOOL ALGEBRA. 


172. Problems involving Fractional Equations. 


Ex. The sum of the third and fourth parts of a certain 
number exceeds 3 times the difference of the fifth and sixth 
parts by 29. Find the number. 


Let x = the number. 
Then j “ ; = the sum of its third and fourth parts, 
: “2 ; = the difference of its fifth and sixth parts, 


3(Z = AS 3 times the difference of its fifth and sixth parts, 
5 


: =i ; —3 & — 4 = the given excess. 
But 29 = the given excess. 
L x 2 
oi eh en ata Peete re | me 
ee! & a) 


Multiply by 60 the L.C.D. of the fractions. 
20” + 152% — 36a + 302 = 60 x 29. 
Combining, 29 2 = 60 x 29. 
“. & = 60, 


Exercise 56. 


1. The sum of the sixth and seventh parts of a number 
is 18. Find the number. 


2. The sum of the third, fourth, and sixth parts of a 
number is 18. Find the number. 


3. The difference between the third and fifth parts of 
a number is 4. Find the number. 


4. The sum of the third, fourth, and fifth parts of a 
number exceeds the half of the number by 17. ‘Find the 
number. 


5. There are two consecutive numbers, 2 and x+1, 
such that one-fourth of the smaller exceeds one-ninth of the 
larger by 11. Find the numbers. 


FRACTIONAL EQUATIONS. 151 


6. Find three consecutive numbers such that if they are 
divided by 7, 10, and 17, respectively, the sum of the quo- 
tients will be 15. 


7. Find a number such that the sum of its sixth and 
ninth parts shall exceed the difference of its ninth and 
twelfth parts by 9. 


8. The sum of two numbers is 91, and if the greater is 
divided by the less the quotient is 2, and the remainder 
is 1. Find the numbers. 


Dividend — Remainder 


= Quotient. 
Divisor Q 


Hint. 

9. The difference of two numbers is 40, and if the greater 

is divided by the less the quotient is 4, and the remainder 
4. Find the numbers. 


10. Divide the number 240 into two parts such that the 
smaller part is contained in the larger part 5 times, with a 
remainder of 6. 


11. If a mixture of alcohol and water the alcohol was 
24 gallons more than half the mixture, and the water was 
4 gallons less than one-fourth the mixture. How many 
gallons were there of each? 


12. The width of a room is three-fourths of its length. 
If the width was 4 feet more and the length 4 feet less, the 
room would be square. Find its dimensions. 


Ex. Hight years ago a boy was one-fourth as old as he 
will be one,year hence. How old is he now? 


Let x = the number of years old he is now. 
Then «— 8 =the number of years old he was eight years ago, 
and x +1 =the number of years old he will be one year hence. 
. ¢—-8=}(rc+1). 


Solving, e=11. 


152 SCHOOL ALGEBRA. 


13. A is 60 years old, and B is three-fourths as old. 
How many years since B was just one-half as old as A? 


14. A father is 50 years old, and his son is half that 
age. How many years ago was the father two and one- 
fourth times as old as his son? ~ 


15. A father is 40 years old, and his son is one-third 
that age. In how many years will the son be half as old 
as his father ? 


Ex. A can do a piece of work in 6 days, and B can do 
it in 7 days. How long will it take both together to do 
the work ? 


Let x = the number of days it will take both together. 
= the part both together can do in one day, 


1 
x 
4 = the part A can do in one day, 
+ =the part B can do in one day, 
1 
7 


and i-++4=the part both together can do in one day. 
Ilex 
Lye Oe 
Te+6a=—42 
13 2 = 42 
z= 335. 


Therefore they together can do the work in 3,3; days. 


16. Aan do a piece of work in 3 days, B in 4 days, and 
Cin 6 days. How long will it take them to do it working 
together ? , | 


17. A can do a piece of work in 23 days, B in 3% days, 
and C in 42 days. How long will it take them to do it 
working together ? 


18. A and B can separately do a piece of work in 12 
days and 20 days, and with the help of C they can do it in 
6 days. How long would it take C to do the work? 


FRACTIONAL EQUATIONS. LS 


19. A and B together can do a piece of work in 10 days, 
A and C in 12 days, and A by himself in 18 days. How 
many days will it take B and C together to do the work? 
How many days will it take A, B, and C together ? 


20. A and B can do a piece of work in 10 days, A and 
Cin 12 days, B and Cin 15 days. How long will it take 
them to do the work if they all work together ? 


21. A cistern can be filled by three pipes in 15, 20, and 
30 minutes respectively. In what time will it be filled if 
the pipes are all running together ? 


22. A cistern can be filled by two pipes in 25 minutes 
and 30 minutes, respectively, and emptied by a third in 20 
minutes. In what time will it be filled if all three pipes 
are running together ? 


23. A tank can be filled by three pipes in 1 hour and 20 
minutes, 2 hours and 20 minutes, and 3 hours and 20 min- 
utes, respectively. In how many minutes can it be filled 
by all three together ? 


Ex. A courier who travels 6 miles an hour is followed, 
after 5 hours, by a second courier who travels 74 miles an 
hour. In how many hours will the second courier overtake 
the first ? 


Let x = the number of hours the first travels. 
Then x — 5 =the number of hours the second travels, 
6a =the number of miles the first travels, 
and («#—5)74=the number of miles the second travels. 
They both travel the same distance. 


in O@i== (#2 —.5).74; 
or 12¢=152— 5. 
x = 25. 


Therefore the second courier will overtake the first in 20 
hours, 


' 


154 SCHOOL ALGEBRA. 


24. A sets out and travels at the rate of 9 miles in 2 
hours. Seven hours afterwards B sets out from the same 
place and travels in the same direction at the rate of 5. 
miles an hour. In how many hours will B overtake A? 


25. A man walks to the top of a mountain at the rate of 
2% miles an hour, and down the same way at the rate of 4 
miles an hour, and is out 5 hours. How far is it to the top 
of the mountain ? 


26. In going from Boston to Portland, a passenger train, 
at 27 miles an hour, occupies 2 hours less time than a freight 
train at 18 miles an hour. Find the distance from Boston 
to Portland. 

27. A person has 6 hours at his disposal. How far may 
he ride at 6 miles an hour so as to return in time, walking 
back at the rate of 3 miles an hour? 

28. A boy starts from Exeter and walks towards Ando- 
ver at the rate of 3 miles an hour, and 2 hours later another 
boys starts from Andover and walks towards Exeter at the 
rate of 24 miles an hour. The distance from Exeter to 
Andover is 28 miles. How far from Exeter will they meet? 


Ex. A hare takes 5 leaps while a greyhound takes 8, but 
1 of the greyhound’s leaps is equal to 2 of the hare’s. The 
hare has a start of 50 of her own leaps. How many leaps 
must the greyhound take to catch her? 

Let 3x =the number of leaps the greyhound takes. 

Then 5a=the number of leaps the hare takes in the same time. 

Also, let a= the number of feet in one leap of the hare. 

Then 2a=the number of feet in one leap of the hound. 

Hence 3x2 xX 2a or 6axr =the whole distance, 
and (5% +50)a or 5ax + 50 =the whole distance. 

.6ax = Sax + 50a. 
xe = 00, 

and 3a = 150. 

Therefore the greyhound must take 150 leaps. 


FRACTIONAL EQUATIONS. 155 


29. A hare takes 7 leaps while a dog takes 5, and 5 of 
the dog’s leaps are equal to 8 of the hare’s. The hare has 
a start of 50 of her own leaps. How many leaps will the 
hare take before she is caught ? 


30. A dog makes 4 leaps while a hare makes 5, but 3 of 
the dog’s leaps are equal to 4 of the hare’s. The hare has a 
start of 60 of the dog’s leaps. How many leaps will each 
take before the hare is caught? | 


Notr. If the number of units in the breadth and length of a 
rectangle are represented by x and x +a, respectively, then «x(x + a) 
will represent the number of units of area in the rectangle. 


31. A rectangle whose length is 23 times its breadth 
would have its area increased by 60 square feet if its length 
and breadth were each 5 feet more. Find its dimensions. 


32. A rectangle has its length 4 feet longer and its width 
3 feet shorter than the side of the equivalent square. Find 
its area. 


33. The width of a rectangle is an inch more than half 
its length, and if a strip 5 inches wide is taken off from the 
four sides, the area of the strip is 510 square inches. Find 
the dimensions of the rectangle. 


Nore. If « pounds of metal lose 1 pound when weighed in 


water, 1 pound of metal will lose ! ofa pound. 
a 


34. If 1 pound of tin loses ,% of a pound, and 1 pound 
of lead loses %, of a pound, when weighed in water, how 
many pounds of tin and of lead in a mass of 60 pounds that 
loses 7 pounds when weighed in water? 


35. If 19 ounces of gold lose 1 ounce, and 10 ounces of 
silver lose 1 ounce, when weighed in water, how many 
ounces of gold and of silver in a mass of gold and silver 
weighing 530 ounces in air and 495 ounces in water? 


156 SCHOOL ALGEBRA. 


Ex. Find the time between 2 and 8 o'clock when the 
hands of a clock are together. 


At 2 o'clock the hour-hand is 10 minute-spaces ahead of the 
minute-hand. 
Let x = the number of spaces the minute-hand moves over. 
Then 2—10= the number of spaces the hour-hand moves over. 
Now, as the minute-hand moves 12 times as fast as the hour-hand, 
12(% — 10) =the number of spaces the minute-hand moves over. 


“. © =12(x%—10), 
and lla = 120. 
*, ©= 1019. 


Therefore the time is 1012 minutes past 2 o’clock. 


36. At what time between 2 and 8 o'clock are the hands 
of a watch at right angles? 


37. At what time between 8 and 4 o'clock are the hands 
of a watch pointing in opposite directions? 


38. At what time between 7 and 8 o'clock are the hands 
of a watch together ? 


Ex. A merchant adds yearly to his capital one-third of 
it, but takes from it, at the end of each year, $5000 for 
expenses. At the end of the third year, after deducting 
the last $5000, he has twice his original capital. How 
much had he at first? 


Let x = number of dollars he had at first. 

Then ** — 5000, or £2=18000 
will stand for the number of dollars at the end of first year, 
Th ! (= aaa — 6000, or 162 a 


will stand for the number of dollars at the end of second year, 
and 4 ie _ eee — 5000, or 64% — 555000 

3 9 27 
will stand for the number of dollars at the end of third year. 


? 


FRACTIONAL EQUATIONS. 157 


But 2x stands for the number of dollars at the end of third year. 
. 64a — 555000 | 
OT We is 

Whence x = 55,500. 


20. 


39. A trader adds yearly to his capital one-fourth of it, 
but takes from it, at the end of each year, $800 for ex- 
penses. At the end of the third year, after deducting the 
last $800, he has 1% times his original capital. How much 
had he at first ? 


40. A trader adds yearly to his capital one-fifth of it, 
but takes from it, at the end of each year, $2500 for ex- 
penses. At the end of the third year, after deducting 
the last $2500, he has 154 times his original capital. 
Find his original capital. 


41. A’sage now is two-fifths of B’s. Eight years ago A’s 
age was two-ninths of B’s. Find their ages. 


42. A had five times as much money as B. He gave B 
5 dollars, and then had only twice as much as B. How 
much had each at first ? 


43. At what time between 12 and 1 o’clock are the hour 
and minute hands pointing in opposite directions ? 


44. Hleven-sixteenths of a certain principal was at in- 
terest at 5 per cent, and the balance at 4 per cent. The 
entire income was $1500. Find the principal. 


45. A train which travels 36 miles an hour is 3 of an 
hour in advance of a second train which travels 42 miles 
an hour. In how long a time will the last train overtake 
the first ? 


46. An express train which travels 40 miles an hour 
starts from a certain place 50 minutes after a freight train, 
and overtakes the freight train in 2 hours and 5 minutes. 
Find the rate per hour of the freight train. 


158 SCHOOL ALGEBRA. 


47. A messenger starts to carry a despatch, and 5 hours 
after a second messenger sets out to overtake the first in 8 
hours. In order to do this, he is obliged to travel 24 miles 
an hour more than the first. How many miles an hour 
does the first travel ? 


48. The fore and hind wheels of a carriage are respec- 
tively 94 feet and 11 feet in circumference. What distance 
will the carriage have made when one of the fore wheels 
has made 160 revolutions more than one of the hind 
wheels ? 


49. When a certain brigade of troops is formed in a 
solid square there is found to be 100 men over; but when 
formed in column with 5 men more in front and 38 men less 
in depth than before, the column needs 5 men to complete 
it. Find the number of troops. 


50. An officer can form his men in a hollow square 14 
deep. The whole number of men is 3136. Find the num- 
ber of men in the front of the hollow square. 


51. A trader increases his capital each year by one- 
fourth of it, and at the end of each year takes out $2400 
for expenses. At the end of 38 years, after deducting the 
last $2400, he finds his capital to be $10,000. Find his 


original capital. 


52. A and B together can do a piece of work in 14 days, 
A and © together in 1$ days, and B and C together in 1% 
days. How many days will it take each alone to do the 
work ? 


53. A fox pursued by a hound has a start of 100 of her 
leaps. The fox makes 3 leaps while the hound makes 2; 
but 3 leaps of the hound are equivalent to 5 of the fox. 
How many leaps will each take before the hound catches 
the fox ? 


FRACTIONAL EQUATIONS. 159 


178. Formulas and Rules. When the given numbers of a 
problem are represented by letters, the result obtained from 
solving the problem is a general expression which includes 
all problems of that class. Such an expression is called a 
formula, and the translation of this formula into words is 
called a rule, 


We will illustrate by examples : 


(1) The sum of two numbers is s, and their difference d; 
find the numbers. 


+ 














Let x = the smaller number ; 
then — a +d =the larger number. 
Hence x+x+ds=s, 
or 24 =s—d. 
Set oe d 
2 
and peda Eee 
_std 
2 
Therefore the numbers are = : aq and = es z 


As these formulas hold true whatever numbers s and d 
stand for, we have the general rule for finding two numbers 
when their sum and difference are given: 


Add the difference to the sum and take half the result for 
the greater number. 

Subtract the difference from the sum and take half the 
result for the smaller number. 


(2) If A can do a piece of work in a days, and B can 
do the same work in 0 days, in how many days can both 
together do it? 


160 SCHOOL ALGEBRA. 


Let ‘ 





a = the required number of days. 
Then, 1 _ the part both together can do in one day. 
x 
Now 1 _ the part A can do in one day, 
a 
and ; = the part B can do in one day ; 
therefore : + ; — the part both together can do in one day. 
a 
NG ee 
a45.D Gis 
ab 
Whence em PEA 


The translation of this formula gives the following rule 
for finding the time required by two agents together to 
produce a given result, when the time required by each 
agent separately is known : 


Duwide the product of the numbers which express the units 
of tume required by each to do the work by the sum of these 
numbers ; the quotient is the tume required by both together. 


174, Interest Formulas. The elements involved in com- 
putation of interest are the principal, rate, time, interest, 
and amount. 


Let =the principal, 
r = the interest of $1 for 1 year, at the given rate, 
¢ = the time expressed in years, 
2 = the interest for the given time and rate, 
a = the amount (sum of principal and interest). 


175. Given the Principal, Rate, and Time; to find the Interest. 


Since 7 is the interest of $1 for 1 year, pr is the interest 
of $p for 1 year, and prt is the interest of $p for ¢ years. 


‘1 pr. (Formula 1.) 


Rue. Multiply together the principal, rate, and time. 


FRACTIONAL EQUATIONS. 161 


176. Given the Principal, Rate, and Time; to find the Amount. 
Since the amount a is the sum of the principal and 
interest, 
a==p- prt. (Formula 2.) 
177. Given the Amount, Rate, and Time; to find the Principal. 
From formula 2, ptprt=a, 
or pitri)=a. 


Divide by 1+ rt, P=j mE cs (Formula 3.) 
r 





178, Given the Amount, Principal, and Rate; to find the Time. 
From formula2, p-+pri=a. 
Transpose p, pri=a—p. 


Divide by pr, t= sate (Formula 4.) 


179. Given the Amount, Principal, and Time; to find the Rate. 
From formula2, p+prt=a. 
Transpose p, prt=a—p. 


Divide by pt, y carn (Formula 5.) 


Exercise 57. 


Solve by the preceding formulas: 


1. The sum of two numbers is 40, and their difference is 
10. Find the numbers. 


2. The sum of two angles is 100°, and their difference is 
21° 30’. Find the angles. 

3. The sum of two angles is 116° 24! 80", and their 
difference is 56° 21! 44", Find the angles. 


162 SCHOOL ALGEBRA. 


4. A can do a piece of work in 6 days, and B in 5 days. 
How long will it take both together to do it? 


8. Find the interest of $2750 for 3 years at 4% per 

cent. 

6. Find the interest of $950 for 2 years 6 months at 5 
per cent. 

7. Find the amount of $2000 for 7 years 4 months 
at 6 per cent. 

8. Find the rate if the interest on $680 for 7 months is 
$35.70. 


9. Find the rate if the amount of $750 for 4 years is 
$900. 


10. Find the rate if a sum of money doubles in 16 years 
and 8 months. 


11. Find the time required for the interest on $2130 to 
be $456.65 at 6 per cent. 


12. Find the time required for the interest on a sum of 
money to be equal to the principal at 5 per cent. 


13. Find the principal that will produce $161.25 interest 


in 8 years 9 months at 8 per cent. 


14. Find the principal that will amount to $1500 in 3 
years 4 months at 6 per cent. 


15. How much money is required to yield $ 2000 interest 
annually if the money is invested at 5 per cent? 


16. Find the time in which $640 will amount to $1000 
at 6 per cent. 


17. Find the principal that will produce $100 per month, 


at 6 per cent. 


18. Find the rate if the interest on $700 for 10 months 
is $25, 


i — 


CHAPTER XI. 


SIMULTANEOUS EQUATIONS OF THE FIRST 
DEGREE. 


180. If we have two unknown numbers and but one rela- 
tion between them, we can find an unlimited number of 
pairs of values for which the given relation will hold true. 
Thus, if z and y are unknown, and we have given only the 
one relation z+ y=10, we can asswme any value for z, 
and then from the relation + y= 10 find the correspond- 
ing value of y. For from 2+y=10 we find y=10—z2. 
If x stands for 1, y stands for 9; if x stands for 2, y stands 
for 8; if x stands for — 2, y stands for 12; and so on with- 
out end. 


181, We may, however, have two equations that express 
different relations between the two unknowns. Such equa- 
tions are called independent equations. Thus, «+ y¥y=10 
and «—y=2 are independent equations, for they ey 
express different relations between x and y. 


182. Independent equations involving the same unknowns 
are called simultaneous equations, 

If we have two unknowns, and have given two independ- 
ent equations involving them, there is but one pair of values 
which will hold true for both equations. Thus, if in § 181, 
besides the relation z+ y= 10, we have also the relation 
x — y = 2, the only pair of values for which both equations 
will hold true is the pair = 6, y= 4. 

Observe that in this problem wz stands for the same num- 
ber in both equations; so also does y. 


164 SCHOOL ALGEBRA. 


183. Simultaneous equations are solved by combining 
the equations so as to obtain a single equation with one 
unknown number; this process is called elimination. 

There are three methods of elimination in general use: 


I. By Addition or Subtraction. 
II. By Substitution. 
III. By Comparison. 


184, Elimination by Addition or Subtraction. 


(1) Solve: 5a—3y= a} (1) 
22+ 5y = 39 (2) 
Multiply (1) by 5, and (2) by 3, 
252 —15y = 100 (3) 
62+ 15y=117 Sale: 
Add (3) and (4), 31a = 217 
. o= 7. 
Substitute the value of « in (2), 
14+ 5y =39, 
“y=, 


In this solution y is eliminated by addition. 


(2) Solve: 62+ 385y=177 (1) 
82—2ly= 33 (2) 
Multiply (1) by 4, and (2) by 3, 
24a + 140y = 708 (3) 
244— 63y= 99 (4) 
Subtract, 203 y = 609 
“ ¥ =3. 
Substitute the value of y in (2), 
8% —63 =33. - 
* e=12. 


In this solution 2 is eliminated by swbtraction. 


SIMULTANEOUS EQUATIONS. 165 


185. Hence, to eliminate by addition or subtraction, we 
have the following rule: 


Multiply the equations by such numbers as will make the 
coefficients of one of the unknown numbers equal in the 
resulting equations. 

Add the resulting equations, or subtract one from the other, 
according as these equal coefficients have unlike or like signs. 

Nors. It is generally best to select the letter to be eliminated 
which requires the smallest multipliers to make its coefficients equal ; 
and the smallest multiplher for each equation is found by dividing 
the L.C.M. of the coefficients of this letter by the given coefficient in 
that equation. Thus, in example (2), the L.C.M. of 6 and 8 (the co- 


efficients of x) is 24, and hence the smallest multipliers of the two 
equations are 4 and 3 respectively. 


Sometimes the solution is simplified by first adding the 
given equations, or by subtracting one from the other. 


(3) 2+49y= 51 (1) 
49x+ y= 99 (2) 
Add (1) and (2), 502 + 50y = 150 (3) 
Divide (3) by 50, e+y =3. (4) 
Subtract (4) from (1), 48 y = 48. 
“y=. 
Subtract (4) from (2), 48% = 96. 
“ = 2, 


Exercise 58. 


Solve by addition or subtraction : 


if > ema 4. cee atc at 
22— y= 3 32—6y=15 
2. aad was? 5. Sa Yi Bo 
- 2a—4y= 4 3842—2y=17 
3. "Sele 6. ete cantet 
x+by= 28 22 —d3y= 26 


166 SCHOOL ALGEBRA. 
Vs pease Lis te? Qe - 
Sly = 8 38%—- 8y=90 
8. ek See f 12. 4¢%— asi 
22+ dy = 43 8x2— 4y=17 
ay Sone mua 13. Tx— ae 
S42 —2y=15 2x —l0y= 39 
10. 24+ y= it 14, 82+ Lae 
Ta+5y=21 22+ 5y=13 


186, Elimination by Substitution. 


(1) Solve: Ha ae 
4z+3y=25 
5a +4y =32 (1) 
4a +3y = 25 (2) 
Transpose 4 y in (1), 5a=32—4y. (3) 
Divide by coefficient of 2, Pyles od (4) 


5 


Substitute the value of a in (2), 


128 —16y + 15y =125, 
—y=-—83. 
ap eee 
Substitute the value of y in (2), 
42 +9 = 25. 
Ye 


Hence, to eliminate by substitution, 


From one of the equations obtain the value of one of the 
unknown numbers in terms of the other. 

Substitute for this unknown number its value in the other 
equation, and reduce the resulting equation. 


° 


SIMULTANEOUS EQUATIONS. 167 


Exercise 59. 


Solve by substitution : 


Cy Ey \ 8. 382— ised 
sa— Sy=11 2z+ dSy=—68 
oye = hy = at 9. 2x— eae 
oz— 2y=—10 52+ 2y= 29 
3. 2Ze— 3y= } 10. 6x%— peiaw 
82— 2Zy= 29 oz— Oy= 8 
Ae et aise Vii Sti test 
22+ Ty=88 2a+ S5y=82 
Sos a = >} 12. «2+ pte 
x+ 2y=25 or+ Zy = 46 
6. peers” = S| 13. 82— Lelia 
l3a— 5y=21 4%x— 3y= 9 
fe ie! 14. 54+ athe ae 
if — Ay 7 82+lly= 1 
187. Elimination by Comparison. 
Solve: pate caeat 
84+ 2y = 23 
2%—5y = 66. (1) 
35 4+ 2y = 23. (2) 
Transpose 5y in (1), and 2y in (2), 
2x = 66 +5y, (3) 
3a” = 23 —2y. (4) 
Divide (3) by 2, pie a (5) 
Divide (4) by 3, Se eh aet, (6) 


Equate the values of 2, = epee (7) 
vo 


168 


SCHOOL ALGEBRA. 


Reduce (7), 


198 + 15y = 46 —4y, 
19y =— 152. 
. y=—8. 


Substitute the value of y in (1), 
2x + 40 = 66. 
“. = 13, 


188. Hence, to eliminate by comparison, 


From each equation obtain the value of one of the unknown 


numbers in terms of the other. 


Form an equation from these equal values and reduce the 


equation. 


Exercise 6O. 


Solve by comparison : 


by 


so 


x + farce 
8%2— 2y=25 
Tx+ Cae 
or— 4y= 
9ua+ se 
4r+ 9y=89 
Tx+ as 
824—-—  y¥= 
ee ee 
38x2--47y = 46 
247—- y= ie 
5a— 38y=14 
llz— Ty= att 
92— 5y=10 
ee Aan 
138%—2y= 57 


15. 


16. 


24— 8y= a 
5a+ 2y=126 
50z— Iy= t 
2 y= BD 

2+2ly= ae 
2iy-+- 2%= 19 
10a + a | 
82+ 10y=125 
62—138y= i 
52—1l2y=. 4 
Qa + Yen 
10%+ 2y= 60 
82— 5y= i 
Ta+ y=265 
12”%+ her 
38y—1l9r%= 3 


SIMULTANEOUS EQUATIONS 169 


189. Each equation must be simplified, if necessary, 
before the elimination. 


Solve: $a—L(iyt+ seit 
Mo+1+iy-1)=9 | 
e—3(yt1)=1. | (1) 
3(@+1)+ey—1)=9. (2) 
Multiply (1) by 4, and (2) by 12, 
34a—2y—2=4, (3) 
4%7+4+9y—9= 168. (4) 
From (3), 3a—2y=6. (5) 
From (4), 44+ 9y=118. (6) 
Multiply (5) by 4, and (6) by 3, 
12%— 8y= 24 
122 4+ 27y = 339 
35y = 315 
 ¥=9. 
Substitute value of y in (1), x = 8, 


Exercise 61. 




















Solve 
Bee. 4.) ae aed ae 
fie ice = 4 ey 
315 3 | 8 1 5 
x pen eee gi OL 3 a 
a28 6. ae J 
eee Fg |, Pe by gs ee ry) 
3 y 9, 5 
oy at+y_p¢ g ta 2y_ rt yy | 
Pegi: 1 | Rec pin | 
40 zy 
22—Y 1 9 ons Die | 
[og Sst cal 


170 


SCHOOL ALGEBRA. 








12-422 — SY by _ Ae 


easy a ed ee e+2y+1l_o 
3 a 5 2z—y+1 
e210.) 0, ares Dee | 38z—y+1_, 
4 3 2 J z—yt+3 
9 aA 10> sya 
ae5h 8 4 
AYP th eae yen 
3 8 
10. tery hs Oa eye 
3 4 
da—4y+3 _ 2y—4e+421 
4 3 
11. UR ee ee | 
5 4 
oy—it , 4@—8 | 
5 + —__ 5 = 18— ae 
1B ig ae ee OY eee | 


Pai eC ee Ma ieeee 





2 6 
2r-+by Je— Ti lyse 
2 8 16 


14, ee 2 


ne at 
3 19 


SIMULTANEOUS EQUATIONS. 171 











15, SEY _? ig, Yo 4 et! | 





5a —6 
shee ere wa) 


ESO et eka 
10 3 ie! 


OYE Leen ey Pe OD ) 
5 30 | 


y—l1 zt 10z—3y—20_ 84+2y+3 
3 20 30 


Norg. -In solving the following problems proceed as in @ 171. 


ro ou te fe ey $3 ty — 4) 


18. 


8 4%—2y 12 
S2+3 (2—dy _62—1 
4 7—2 3 


ey ee 
20. ¢— =r — 9 


93 — a 4 


— 3 
Gemini 





fer7, de—4y _lit 8% 


1. 
: 3 2x2+8 6 
Pie pao 6y-13. 10a 5384 
4 2x—s8y 8 J 
poem Sle 2y) ise) 
ci Bale. 2(x — 4) 5 


eer eo Tee i ok By+ 5a 
4 4 2(2y— 8) 


72 SCHOOL ALGEBRA. 


190, Literal Simultaneous Equations. 

Solve: ax + by =e } 

az + bly =c' 

Note. The letters a’, b’ are read a prime, b prime. In like man- 
ner, a/’, a//’ are read a second, a third, and ay, dy, ds are read @ sub 
one, a sub two, a sub three. It is sometimes convenient to represent 
different numbers that have a common property by the same letter 


marked by accents or suffives. Here a and a’ have a common 
property as coefficients of a. 


av +by =c. (1) 
a/c + b/y=c?’. (2) 
To find the value of y, multiply (1) by a’, and (2) by a, 
aa’x + a/by =a/e | 
aa’x + ab/y = ac’ 


a/by — ab’y = a/c — ac’ 


To find the value of x, multiply (1) by 0’, and (2) by 5. 
ab’ + bb’y = b/c 
a/ba + bb’/y = be’ 
ab/a — a/bu = b/c — be’ 

__ b/c — be? 


H 4; ee 
ab/—a/b 


Exercise 62. 


Solve: 

i Diem) 5. eae 
zr—y=d x= dy 

2. mx Meee 6. bx bbsas, 
ma + n'y =r! b'x—aly=1 

3. Si are ie pee 
ale + bly =! 4bx— 8ay= tab 

4, «£— eit 8. 22 eae, 
cu + aby = ms 3a —2y=a+b 


ee 


SIMULTANEOUS EQUATIONS. LS 





























bx Ce, ek 
cing ra high Gath 
bz +cy=a+b 
y—n db bz +ay=c 
b = 
he o4 15 8a? + samen es 
ie ott | ax+2by=d 
BO 
2r yl geen ae celle Ws Liste) 
ab 3 | Oy a eee 
1 
12. 2 Yorn. | ane 
pegs) gi 5 a+b a—b a-—6b| 
2+y=2a 


x Yan oe 
PE fe ‘ a | 
13. an, ty _3y 4) ape 


ee 








pei ey n'y 
x—y=a—b n—a n—b 
191, Fractional simultaneous equations, of which the de- 


nominators are simple expressions and contain the unknown 
numbers, may be solved as follows: 


(1) Solve: oo =m | 
YT, 
aa | 
— + —-—7N 
oe RD 
We have ed aa (1) 
way 
and e+ San, (2) 
To find the value of y. 
Multiply (1) by «¢, = + = = cm, (3) 


174 SCHOOL ALGEBRA. 
Multiply (2) by a, weds aos an. (4) 
x y 


Subtract (4) from (3), bc — ad 





= cm— an, 


y 
Multiply both sides by y, be —- ad =(em — an)y. 











hs: be — ad 
°F om — an 
To find the value of x. 
ad bd 
Multiply (1) by d, — +—=dm. (5) 
wy 
Multiply (2) by 3, be + bigs bn. (8) 
J es, 
Subtract (6) from (5), nacelle bn. 
v 
Multiply both sides by a, ad — be = (dm — bn)z. 
_ ad—be 
dm — bn 
4) 2 
2) Solve: ea Backes = hy 
(2) ate 
ace Se Rs a 
627 10 
We have =. + = noite (1) 
7 1 
ant AS oe 
a0 6x 10y (?) 
Multiply (1) by 15, the L.C.M. of 3 and 5, and (2) by 30, 
oP EN TOG: (3) 
| 
35 3 
sei Pb a 4 
ney 90. (4) 
Multiply (4) by 2, and add the result to (3), 
2 _ 286, 
ae 
ere 
: 3 
Substitute the value of a in (1), and we get 
1 
ere 


SIMULTANEOUS EQUATIONS. 


Exercise 63. 


Bg se x | et 
ae. | 7 ee eee 
leah 
weersoy | Esthet 
Dies 2h) ee fe 
EEA 8 Tela, | 
y zy 
ro nm ol 
2208 eee | 
y aly J 
a) oe mn 
y 3 | 9. miyr’ | 
Y aly 
Cf ee (ty 40 4 
3 co aatk 
ee a7 ale eet ipa 
y coy a) 
5 oun 
eee an 1) 11. ———= 
wee aa | Ct bY | 
ah Se Re Bot 
BUM, ax a= 04) J 
Bere | [ome ere arene 
3Yy OL Ly 
4 Sere 
pe — — == GQ" 3 
is, ia ath 


175 


176 SCHOOL ALGEBRA. 


192. If three simultaneous equations are given, involy- 
ing three unknown numbers, one of the unknowns must be 
eliminated between two pavrs of the equations; then a 
second unknown between the two resulting equations. 

Likewise, if four or more equations are given, involving 
four or more unknown numbers, one of the unknowns must 
be eliminated between three or more pairs of the equations ; 
then a second between the pairs that can be formed of the 
resulting equations; and so on. 

Norr. The pairs chosen to eliminate from must be independent 


pairs, so that each of the given equations shall be used in the process 
of the eliminations. 


Solve: 2%—8y+4z2= 4 (1) 
Seb 5y—Te=12| (2) 
S2— y—8z2= 5 (3) 
Eliminate z between the equations (1) and (8). 


Multiply (1) by 2, 4x—6y+8z= 8 (4) 
(3) is 5e— y—8z2= 5 
Add, 9x—Ty = 13 (5) 


Eliminate z between the equations (1) and (2). 

Multiply (1) by 7, 14”%—21ly + 282 = 28 

Multiply (2) by 4, 124+ 20y—28z=48 

Add, 26a—- y = 76 (6) 

We now have two equations (5) and (6) involving two unknowns, 
«and y. 


Multiply (6) by 7, 182% -— Ty = 532 (7) 
(5) is 9a—Ty= 13 
Subtract (5) from (7), 173 2 = 519 

*. i mak, 
Substitute the value of x in (6), 78 — y = 76. 

 y=2. 
Substitute the values of x and y in (1), 

6—6442=4. 


*, =. 


Oe i 


SIMULTANEOUS EQUATIONS. 


177 


Exercise 64, 


~ oty— 8=0 
y+-2—28=0 
a+z2—14=0 


44+ 3y+22= 25 
34% —2y+5z2= 20 
we 


Tot Sy ever 
x— a Zt 


sy a7 0 


. loez— y+3z2=42 
Ta+t2y+ z2=951 
8a+38y— z2= 24 
. 5a+2y—82=—160 
82+9y+82=115 
22—8y—5z= 40 


. 62—2y+5z2=53 
5a+38y+7z2= 33 
e+ yt z= 9d 


62+ 2y—Tz= 5 


| 
oe 
| 
| 
| 


: Se-h2y—Te= | 


22 — 


yt+8z2=45 


. 22+ 7y+10z2=25 
ety— 424=9 
Tx—Ty—11z2=78 


| 


10. 


12. 


13. 


14. 


15. 


18.., 


82—6y+ 7z=51 
4x+-8y— 92= 53 


2+2y+10z2= | 


or oyy reall 


82+3y+t Tz= 3884 


2e+ yt 2=256 


2z2= zxr—d3y—18 


aaa 1 


zx— yt+38z2=13 
l0y+5x2—22=48 


2¢-+ 3y— 4z2=1 
10z2— -6y+12z2= 

xtl2y+ 2z=5 
82+ by+2z2=3 
l2y+ 42—62= 
9¢4+18y—4z2=4 


5y—4x—42=1 
sa+9y+ 2=9 
sat2y+t 2=203 
20-4 Y + 22 = 26h | 
a+ y+10z2=55 


. 2a+ tai 


SCHOOL ALGEBRA. 


178 


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all 
mila wala -. 
6 | = D> 
sian O!1a on 
ee ee 


mary eR ee 
| 


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N SH OD 


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AID AIH Gs!aR 


+ | + 


CQ es 8 CO em rt SS 


16 
RN 
a ee 
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ee eerie 
AIA HIR OLD 
ee a a 
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& 


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sla a8 4/8 
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COlMm lS wtls 
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HIS Ala Hla 


R 
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CHAPTER XIL 


PROBLEMS INVOLVING TWO UNKNOWN 
NUMBERS. 


193. It is often necessary in the solution of problems to 
employ two or more letters to represent the numbers to be 
found. In all cases the conditions must be sufficient to 
give just as many equations as there are unknown numbers 
employed. 


194, If there are more equations than unknown numbers, 
some of them are superfluous or inconsistent; if there are 
less equations than unknown numbers, the problem is inde- 
terminate. 


(1) If A gives B $10, B will have three times as much 
money as A. If B gives A $10, A will have twice as much 
money as B. How much has each? 


Let x = number of dollars A has, 
and y = number of dollars B has. 


Then, after A gives B $10, 
x — 10 =the number of dollars A has, 
y + 10 =the number of dollars B has. 
“ ¥ +10 =3(e— 10). (1) 
If B gives A $10, 
x +10=the number of dollars A has, 


y — 10 = the number of dollars B has. 
. ©+10=2(y— 10). (2) 


From the solution of equations (1) and (2), # = 22, and y = 26. 


Therefore A has $22, and B has $26. 


180 SCHOOL ALGEBRA. 


(2) If the smaller of two numbers is divided by the 
greater, the quotient is 0.21, and the remainder 0.0057; 
but if the greater is divided by the smaller, the quotient 
is 4 and the remainder 0.742. Find the numbers. 


Let « = the greater number, 
and y = the smaller number. 
Then Ye OOO oe (1) 
2 
and %— 0.742 _ 4. (2) 
_ 
“. y — 0.21 4% = 0.0057, (3) 
e x—4y =0.742. (4) 
Multiply (3) by 4, 4 —0.84.¢ = 0.0228 (5) 
(4) is —4y+ x = 0.742 
By adding, 0.16 2 = 0.7648 
eel oy 
Substituting the value of z in (4), 
—4y = — 4.088, 
“. y = 1.0095, 


Exercise 65. 


1. If A gives B $100, A will then have half as much 
money as B; but if B gives A $100, B will have one-third 
as much as A. How much has each? 


2. If the greater of two numbers is divided by the 
smaller, the quotient is 4 and the remainder 0.37; but if 
the smaller is divided by the greater, the quotient is 0.23 
and the remainder 0.0149. Find the numbers. 


3. A certain number of persons paid a bill. If there 
had been 10 persons more, each would have paid $2 less; 
but if there had been 5 persons less, each would have paid 
$2.50 more. Find the number of persons and the amount 
of the bill. 


a a oe ae 


PROBLEMS. 181 


4, A train proceeded a certain distance at a uniform 
rate. Ifthe speed had been 6 miles an hour more, the time 
occupied would have been 5 hours less; but if the speed had 
been 6 miles an hour less, the time occupied would have 
been 74 hours more. Find the distance. 


Hint. If «=the number of hours the train travels, and y the 
number of miles per hour, then xy = the distance. 


5. A man bought 10 cows and 50 sheep for $750. He 
sold the cows at a profit of 10 per cent, and the sheep at a 
profit of 80 per cent, and received in all $875. Find the 
average cost of a cow and of a sheep. 


6. It is 40 miles from Dover to Portland. A sets out 
from Dover, and B from Portland, at 7 o’clock A.m., to meet 
each other. A walks at the rate of 34 miles an hour, but 
stops 1 hour on the way; B walks at the rate of 24 miles an 
hour. At what time of day and how far from Portland will 
they meet? 


7. The sum of two numbers is 35, and their difference 
exceeds one-fifth of the smaller number by 2. Find the 
numbers. 


8. If the greater of two numbers is divided by the 
smaller, the quotient is 7 and the remainder 4; but if three 
times the greater number is divided by twice the smaller, 
the quotient is 11 and the remainder 4. Find the numbers. 


9. If 3 yards of velvet and 12 yards of silk cost $60, 
and 4 yards of velvet and 5 yards of silk cost $58, what is 
the price of a yard of velvet and of a yard of silk? 


10. If 5 bushels of wheat, 4 of rye, and 3 of oats are sold 
for $9; 3 bushels of wheat, 5 of rye, and 6 of oats for $8.75; 
and 2 bushels of wheat, 3 of rye, and 9 of oats for $7.25; 
what is the price per bushel of each kind of grain? 


182 SCHOOL ALGEBRA. 


Nore I. A fraction the terms of which are unknown may be repre- 


sented by ”. 
y 


Ex. A certain fraction becomes equal to 4 if 2 is added 
to its numerator, and equal to 4 if 3 is added to its denomi- 
nator. Find the fraction. 








Let : = the required fraction. 
a 
9 
Then shililed ie 
y 
and tS 
yts 


The solution of these equations gives 7 for #, and 18 for y. 


Therefore the required fraction is 7%. 


11. A certain fraction becomes equal to 4 if 3 is added 
to its numerator and 1 to its denominator, and equal to 4 
if 3 is subtracted from its numerator and from its denomi- 
nator. Find the fraction. 


12. A certain fraction becomes equal to 5% if 1 is added 
to double its numerator, and equal to 4 if 3 is subtracted 
from its numerator and from its denominator. Find the 
fraction. 


13. Find two fractions with numerators 11 and 5 respec- 
tively, such that their sum is 14, and if their denominators 
are interchanged their sum is 24. 


14. There are two fractions with denominators 20 and 
16 respectively. The fraction formed by taking for a nu- 
merator the sum of the numerators, and for a denominator 
the sum of the denominators, of the given fractions, is equal 
to 4; and the fraction formed by taking for a numerator 
the difference of the numerators, and for a denominator the 
difference of the denominators, of the given fractions, is equal 
to 4. Find the fractions. 


PROBLEMS. 183 


Norse If. A number consisting of two digits which are unknown 
may be represented by 104+, in which x and y represent the digits 
of the number. Likewise, a number consisting of three digits which 
are unknown may be represented by 1002 + 10y +2, in which a, y, 
and z represent the digits of the number. For example, the expres- 
sion 364 means 300 + 60+ 4; or, 100 times 3 + 10 times 6 + 4. 


Ex. The sum of the two digits of a number is 10, and if 
18 is added to the number, the digits will be reversed. 
Find the number. 


Let x = tens’ digit, 
and y = units’ digit. 
Then 10x + y =the number. 
Hence x+y =10, (1) 
and 10a+y+18=10y+2. (2) 
| From (2), 92—-9y=— 18, 
or x—y=—2. (3) 
Add (1) and (8), 22 = 8, 
and therefore ton A, 
Subtract (3) from (1), 2y = 12, 
and therefore y= 6. 


Therefore the number is 46. 


15. The sum of the two digits of a number is 9, and if 27 
is subtracted from the number, the digits will be reversed. 
Find the number. 


16. The sum of the two digits of a number is 9, and if 
the number is divided by the sum of the digits, the quotient 
is 5, Find the number. 


17. A certain number is expressed by two digits. The 
sum of the digits is 11. If the digits are reversed, the new 
number exceeds the given number by 27. Find the number. 


18. A certain number is expressed by three digits. The 
sum of the digits is 21. The sum of the first and last digits 
is twice the middle digit. Ifthe hundreds’ and tens’ digits 
are interchanged, the number is diminished by 90. Find 
the number. 


184 SCHOOL ALGEBRA. 


19. A certain number is expressed by three digits, the 
units’ digit being zero. If the hundreds’ and tens’ digits are 
interchanged, the number is diminished by 180. If the 
hundreds’ digit is halved, and the tens’ and units’ digits are 
interchanged, the number is diminished by 336. Find the 
number. ° 


20. A number is expressed by three digits. If the digits 
are reversed, the new number exceeds the given number by 
99. Ifthe number is divided by nine times the sum of its 
digits, the quotient is 8. The sum of the hundreds’ and 
units’ digits exceeds the tens’ digit by 1. Find the number. 


Nore III. If a boat moves at the rate of x miles an hour in still 
water, and if it is on a stream that runs at the rate of y miles an 


hour, then x +y represents its rate down the stream, 


x —y represents its rate up the stream. 


21. A boatman rows 20 miles down a river and back in 
8 hours. He finds that he can row 5 miles down the river 
in the same time that he rows 3 miles up the river. Find 
the time he was rowing down and up respectively. 


22. A boat’s crew which can pull down a river at the 
rate of 10 miles an hour finds that it takes twice as long to 
row a mile up the river as to row a mile down. Find the 
rate of their rowing in still water and the rate of the 
stream. 


23. A boatman rows down a stream, which runs at the 
rate of 21 miles an hour, for a certain distance in 1 hour 
and 80 minutes; it takes him 4 hours and 80 minutes to 
return. Find the distance he pulled down the stream and 
his rate of rowing in still water. 


Note LV. It is to be remembered that if a certain work can be 
done in # units of time (days, hours, etc.), the part of the work dong 


in one unit of time will be represented by —. 
 ¢ 


i et be 


. oe, 


ee ee 


PROBLEMS. 185 


Ex. A cistern has three pipes, A, B, and C0. A and B 
will fill the cistern in 1 hour and 10 minutes, A and C in 
1 hour and 24 minutes, B and C in 2 hours and 20 minutes. 
How long will it take each pipe alone to fill it? 

1 hour and 10 minutes = 70 minutes. 


1 hour and 24 minutes = 84 minutes. 
2 hours and 20 minutes = 140 minutes. 


Let x = number of minutes it takes A to fill it, 
y = number of minutes it takes B to fill it, 
and z= number of minutes it takes C to fill it. 
Then 1 1 es the parts A, B, and C can fill in one minute 
ma” respectively, 
and : + ; = the part A and B together can fill in one minute. 
But = = the part A and B together can fill in one minute. 
Pe AS 
.-f-=— 1 
geet 70 «) 
In lke manner, 2 “ By (2) 
ep 64 
Lihat Sediag} 
d imi age a Es 3 
on yee HO ®) 
Add, and divide by 2, ee (4) 
By a. BO 
Lis eats bi 
Subtract (1) f 4), Se 
ubtract (1) from (4) Si 
Subtract (2) from (4), fem oe 
y 210 
Subtract (3) from (4), ins Be 
ge. 105 


Therefore x, y, 2 = 105, 210, 420, respectively. 

Hence A can fill it in 1 hour and 45 minutes, B in 3 
hours and 80 minutes, and C in 7 hours. 

24. A and B can do a piece of work together in 8 days, 
A and C in 4 days, B and C in 44 days, How long will it 
take each alone to do the work? 


186 SCHOOL ALGEBRA, 


25. A and B can do a piece of work in 24 days, A and 
C in 34 days, B and © in 4 days. How long will it take 
each alone to do the work? 


26. A and B can do a piece of work in a days, A and O 
in 6 days, B and C ine days. How long will it take each 
alone to do the work ? 


Note V. If represents the number of linear units in the length, 
and y in the width, of a rectangle, xy will represent the number of 
its units of surface; the surface unit having the same name as the 
linear unit of its side. 


27. If the length of a rectangular field were increased 
by 5 yards and its breadth increased by 10 yards, its area 
would be increased by 450 square yards; but if its length 
were increased by 5 yards and its breadth diminished by 
10 yards, its area would be diminished by 350 square yards. 
Find its dimensions. 


28. If the floor of a certain hall had been 2 feet longer 
and 4 feet wider, it would have contained 528 square feet 
more; but if the length and width were each 2 feet less, it 
would contain 316 square feet less. Find its dimensions. 


29. If the length of a rectangle was 4 feet less and the 
width 8 feet more, the figure would be a square of the same 
area as the given rectangle. Find the dimensions of the 
rectangle, 


Nore VI. In considering the rate of increase or decrease in quan- 
tities, it is usual to take 100 as a common standard of reference, so 
that the increase or decrease is calculated for every 100, and there- 
fore called per cent. 

It is to be observed that the representative of the number result- 
ing after an increase has taken place is 100 + increase per cent; and 
after a decrease, 100 — decrease per cent. 

Interest depends upon the time for which the money is lent, as 


PROBLEMS. 187 


well as upon the rate per cent charged; the rate per cent charged 
being the rate per cent on the principal for one year. Hence, 


Simple interest = /tncipal x Rate Ber.cenb x Tho, 


where Time means number of years or fraction of a year. 


Amount = Principal + Interest. 


In questions relating to stocks, 100 is taken as the representative 
of the stock, the price represents its market value, and the per cent 
represents the interest which the stock bears. Thus, if six per cent 
stocks are quoted at 108, the meaning is, that the price of $100 of 
the stock is $108, and that the interest derived from $100 of the 
stock will be ~$,5 of $100, that is, $6 a year. The rate of interest on 
the money invested will be 499 of 6 per cent. 


30. A man has $10,000 invested. For a part of this 
sum he receives 5 per cent interest, and for the rest 6 per 
cent; the income from his 5 per cent investment is $60 
more than from his 6 per cent. How much has he in each 
investment ? 


31. A sum of money, at simple interest, amounted in 4 
years to $ 29,000, and in 5 years to $30,000. Find the sum 
and the rate of interest. 


32. A sum of money, at simple interest, amounted in 10 
months to $2100, and in 18 months to $2180. Find the 
sum and the rate of interest. 


33. A person has a certain capital invested at a certain 
rate per cent. Another person has $2000 more capital, 
and his capital invested at one per cent better than the 
first, and he receives an income of $150 greater. A third 
person has $3000 more capital, and his capital invested at 
two per cent better than the first, and he receives an income 
of $280 greater. Find the capital of each and the rate at 
which it is invested. 


188 SCHOOL ALGEBRA. 


34. A sum of money, at simple interest, amounted in m 
years to c dollars, and in years to d dollars. Find the 
sum and the rate of interest. 


35.. A sum of money, at simple interest, amounted in m 
months to a dollars, and in m months to 6 dollars. Find 
the sum and the rate of interest. 


36. A person has $18,375 to invest. He can buy 3 per 
cent bonds at 75, and 5 per cent bonds at 120. How much 
of his money must he invest in each kind of bonds in order 
to have the same income from each investment? 


Hint. Notice that the 3 per cent bonds at 75 pay 4 per cent on 
the money invested, and 5 per cent bonds at 120 pay 43 per cent. 


37. A man makes an investment at 4 per cent, and a 
second investment at 44 per cent. His income from the 
two investments is $715. If the first investment had been 
at 44 per cent and the second at 4 per cent, his income would 
have been $730. Find the amount of each investment. 


(1) In a mile race A gives B a start of 20 yards and 
beats him by 80 seconds. At the second trial A gives Ba 
start of 32 seconds and beats him by 955 yards. Find the 
number of yards each runs a second. 

Let 2 = number of yards A runs a second, 
and y = number of yards B runs a second. 


Since there are 1760 yards in a mile, 


1760 = number of seconds it takes A to run a mile. 





Since B has a start of 20 yards, he runs 1740 yards the first trial ; 
and as he was 30 seconds longer than A, 


1760 + 30 = the number of seconds B was running. 
x 
But Lidia the number of seconds B was running. 
ALLEL 0 


PROBLEMS. 189 


In the second trial B runs 1760 — 9,3, = 175038 yards. 
_ 17508, _ 1760 
ee y a! x 

From the solution of equations (1) and (2), « = 513, and y= 53. 


+ 82. (2) 


Therefore A runs 532 yards a second, and B runs 5,3, 
yards a second. 


(2) A train, after travelling an hour from A towards B, 
meets with an accident which detains it half an hour; after 
which it proceeds at four-fifths of its usual rate, and arrives 
an hour and a quarter late. If the accident had happened 
80 miles farther on, the train would have been only an hour 
late. Find the usual rate of the train. 


Since the train was detained 4 an hour and arrived 11 hours late, 
the running time was ~ of an hour more than usual. 


Let y = number of miles from A to B, 
and 5x = number of miles the train travels per hour. 
Then y—5a2=number of miles the train has to go after the 
accident. 
Hence ae = number of hours required usually, 
© 
and eae = number of hours actually required. 
© 
‘. Y—S5@_Y—52 _ hogs in hours of running time. 
4a 5a 
But 2 = loss in hours of running time. 
_¥—5e_y—5a_3 (1) 
"Ae 5a 4 


If the accident had happened 30 miles farther on, the remainder 
of the journey would have been y—(5 + 30), and the loss in running 
time would have been 4 an hour. 

_ y—(6e+4+30)_ y—(Gr+30)_1 
- 4% 5x 2 
From the solution of equations (1) and (2), « =6, and 5a = 30. 


(2) 


Therefore the usual rate of the train is 80 miles an hour. 


190 SCHOOL ALGEBRA. 


38.. Two men, A and B, run a mile, and A wins by 2 
seconds. In the second trial B has a start of 184 yards, 
and wins by 1 second. Find the number of yards each 
runs a second, and the number of miles each would run in 
an hour. 


39. Ina mile race A gives B a start of 8 seconds, and is 
beaten by 124 yards. In the second trial A gives B a start 
of 10 yards, and the race is a tie. Find the number of 
yards each runs a second. At this rate, how many miles 
could each run in an hour? 


40. In amile race A gives B a start of 44 yards, and is 
beaten by 1 second. Ina second trial A gives B a start of 
6 seconds, and beats him by 9% yards. Find the number 
of yards each runs a second. 


41. An express train, after travelling an hour from A 
towards B, meets with an accident which delays it 15 min- 
utes. It afterwards proceeds at two-thirds its usual rate, 
and arrives 24 minutes late. If the accident had happened 
5 miles farther on, the train would have been only 21 
minutes late. Find the usual rate of the train. 


42. A train, after running 2 hours from A towards B, 
meets with an accident which delays it 20 minutes. It 
afterwards proceeds at four-fifths its usual rate, and arrives 
1 hour and 40 minutes late. If the accident had happened 
40 miles nearer A, the train would have been 2 hours late. 
Find the usual rate of the train. 


43. A and B can do a piece of work in 24 days, A and 
C in 34 days, B and C in 33 days. In what time can all 
three together, and each one separately, do the work? 


44, A sum of money, at interest, amounts in 8 months 
to $1488, and in 15 months to $1530. Find the principal 
and the rate of interest. 





PROBLEMS. j 191 


45. A number is expressed by two digits, the units’ digit 
being the larger. If the number is divided by the sum of 
its digits, the quotient is 4. If the digits are reversed and 
the resulting number is divided by 2 more than the differ- 
ence of the digits, the quotient is 14. Find the number. 


46. A and B together can dig a well in 10 days. They 
work 4 days, and B finishes the work in 16 days. How 
long would it take each alone to dig the well? 


47. The denominator of the greater of two fractions is 
20. The fraction formed by taking for a numerator the 
sum of the numerators of the two fractions, and for a 
denominator the sum of the denominators, is equal to 2. 
The fraction similarly formed with the difference of the 
numerators, and of the denominators, is equal to 4. The 
sum of the numerators is twice the difference of the denomi- 


nators. Find the fractions. 


48. A cistern can be filled in 5 hours by two pipes, A 
and B, together. Both are left open for 3 hours and 45 
minutes, and then A is shut, and B takes 3 hours and 45 
minutes longer to fill the cistern. How long would it take 
each pipe alone to fill the cistern ? 


49. A man put at interest $20,000 in three sums, the 
first at 5 per cent, the second at 44 per cent, and the 
third at 4 per cent, receiving an income of $905 a year. 
The sum at 44 per cent is one-third as much as the other 
two sums together. Find the three sums. 


50. An income of $335 a year is obtained from two in- 
vestments, one in 44 per cent stock and the other in 5 per 
cent stock. If the 44 per cent stock should be sold at 
110, and the 5 per cent at 125, the sum realized from both 
stocks together would be $8300. How much of each stock 
is there? 


192 SCHOOL ALGEBRA. 


51. A boy bought some apples at 3 for 5 cents, and 
some at 4 for 5 cents, paying for the whole $1. He sold 
them at 2 cents apiece, and cleared 40 cents. How many 


of each kind did he buy ? 


52. Find the area of a rectangular floor, such that if 3 
feet were taken from the length and 3 feet added to the 
breadth, its area would be increased by 6 square feet, but 
if 5 feet were taken from the breadth and 3 feet added to 
the length, its area would be diminished by 90 square feet. 


53. A courier was sent from A to B, a distance of 147 
miles. After 7 hours, a second courier was sent from A, 
who overtook the first just as he was entering B. The time 
required by the first to travel 17 miles added to the time 
required by the second to travel 76 miles is 9 hours and 
40 minutes. How many miles.did each travel per hour? 


54. A box contains a mixture of 6 quarts of oats and 9 
of corn, and another box contains a mixture of 6 quarts of 
oats and 2 of corn. How many quarts must be taken from 
each box in order to have a mixture of 7 quarts, half oats 
and half corn? 


55. A train travelling 30 miles an hour takes 21 minutes 
longer to go from A to B than a train which travels 36 
miles an hour. Find the distance from A to B. 


_56. A man buys 570 oranges, some at 16 for 25 cents, 
and the rest at 18 for 25 cents. He sells them all at the 
rate of 15 for 25 cents, and gains 75 cents. Hew many of 
each kind does he buy ? 


57. A and B run a mile race. In the first heat B 
receives 12 seconds start, and is beaten by 44 yards. In 
the second heat B receives 165 yards start, and arrives at 
the winning post 10 seconds before A. Find the time in 
which each can run a mile, 


PROBLEMS. 1938 


INDETERMINATE PROBLEMS. 


195. If a sengle equation is given which contains éwo 
unknown numbers, and no other condition is imposed, the 
number of its solutions is wnlumited; for, if any value be 
assigned to one of the unknowns, a corresponding value 
may be found for the other. Such an equation is said to 
be indeterminate. 


186. The values of the unknown numbers in an inde- 
terminate equation are dependent upon each other ; so that, 
though they are unlimited in number, they are confined to 
a particular range. 

This range may be still further limited by requiring these 
values to satisfy some given condition; as, for instance, that 
they shall be positive integers. With such restrictions the 
equation may admit of a definite number of solutions. 


Ex. A number is expressed by two digits. If the num- 
ber is divided by the sum of its digits diminished by 4, the 
quotient is 6. Find the number. 


The single statement is 


e+y—-4 
Whence 4a =5y— 24, 
and v=y + 7 _ 6 
; + 
eae 
=y Oa 
We see from that the values of y which will be integral are 4, 8, 
12, 16, or some other multiple of 4, and from the relation s=y—6 44 


4 
that the least positive integral value of y which will give to x a post- 


tive integral value, is 8. If we put 8 for y in (1), we find a=4, Hence 
the number required is 48, 


194 SCHOOL ALGEBRA. 


Exercise 66. 


1. A number is expressed by two digits. If the number 
is divided by the last digit, the quotient is 15. Find the 
number. 


2. A number is expressed by three digits. The sum of 
the digits is 20. If 16 is subtracted from the number and 
the remainder divided by 2, the digits will be reversed. 


Find the number. As 
Here e+y+z2= 20, 

and eee = 1002 ede 
Eliminate y and reduce, and we have 


4%=72+ 8. 


3. A man spends $114 in buying calves at $5 apiece, 
and pigs at $3 apiece. How many did he buy of each? 


4. In how many ways can a man pay a debt of $87 
with five-dollar bills and two-dollar bills? 


5. Find the smallest number which when divided by 5 
or by 7 gives 4 for a remainder. 
n—4 








Let n = the number, then =, and 


m—4 
a 
6. A farmer sells 15 calves, 14 lambs, and 138 pigs for 
$200. Some days after, at the same price, he sells 7 calves, 
11 lambs, and 16 pigs, for which he receives $141. What 


was the price of each? 


Discussion OF PROBLEMS. 


197, The discussion of a problem consists in making 
various suppositions as to the relative values of the given 
numbers, and explaining the results. We will illustrate by 
an example: 


PROBLEMS. 195 


Two couriers were travelling along the same road, and in 
the same direction, from C towards D. A travels at the 
rate of m miles an hour, and B at the rate of m miles an 
hour. At 12 o’clock B was d miles in advance of A. When 
will the couriers be together ? 


Suppose they will be together x hours after 12. Then A has tray- 
elled mz miles, and B has travelled nz miles, and as A has travelled 
d miles more than B 

mz = nx + d, 
or mz — nx = d. 


d 


m—mnN 





_t=> 


Discussion oF THE Prosiem. 1. If m is greater than n, the value 





of x, namely, d , 1s positive, and it is evident that A will over- 
m—n 


take B after 12 o'clock. 





2. If m is less than n, then d_ will be negative. In this case 
—n 


m 
B travels faster than A, and as he isd miles ahead of A at 12 o’clock, 
it is evident that A cannot overtake B after 12 o'clock, but that B 


passed A before 12 o’clock by 


fore, that the couriers were together after 12 o'clock was incorrect, 
and the negative value of x points to an error in the supposition. 


hours. The supposition, there- 








3. If m equals n, then the value of a, that is, , assumes the 


m—n 

form < Now if the couriers were d miles apart at 12 o'clock, and if 
they had been travelling at the same rates, and continue to travel at 
the same rates, it is obvious that they never had poe together, and 


that they never will be together, so that the symbol £ may be regarded 
as the symbol of impossibility. 





4, If m equals n and d is 0, then becomes >. Now if the 


m—n 

couriers were together at 12 o'clock, and if they had been travelling 
at the same rates, and continue to travel at the same rates, it is: 
obvious that they had been together all the time, and that ae will 


continue to be together all the time, so that the symbol 2 5 may be 
regarded as the symbol of indetermination. 


196 SCHOOL ALGEBRA. 


| Exercise 67. 


1. A train travelling 6 miles per hour is m hours in 
advance of a second train which travels a miles per hour. 
In how many hours will the second train overtake the first ? 

, bm 
Ans. 
a—b 
Discuss the result (1) when a>); (2) when a=); (3) when a <b. 





2. A man setting out on a journey drove at the rate of 
a miles an hour to the nearest railway station, distant d 
miles from his house. On arriving at the station he found 
that the train for his place of destination left ¢ hours before. 
At what rate should he have driven in order to reach the 
station just in time for the train? i 


Ans. 


b—ac 
Discuss the result (1) when c=0; (2) when one (3) when 
en—2. In case (2), how many hours did the man have to drive 


from his house to the station? In case (3), what is the meaning of 
the negative value of ¢? 


3. A wine merchant has two kinds of wine which he sells, 
one at a dollars, and the other at d dollars per gallon. He 
wishes to make a mixture of / gallons, which shall cost him 
on the average m dollars a gallon. How many gallons 
must he take of each? 


Ans. ee of the first ; oe of the second. 
a— a 


Discuss the question (1) when a= 6; (2) when a or b=™m; (3) 
when a=b=m; (4) when a>b and <m; (5) when a>6 and 
b>m. 


CHAPTR Ro ALI 
INEQUALITIES. 


198, An inequality consists of two unequal numbers 
connected by the sign of inequality. Thus, 12>4 and 
4 < 12 are inequalities. 


199. Two inequalities are said to be of the same direction 
if the first members are both greater or both less than the 
second members; that is, if the signs of inequality point in 
the same direction. 


200. Two inequalities are said to be the reverse of each 
other if the signs point in opposite directions. 


201. If equal numbers are added to, or subtracted from, 
the members of an inequality, the inequality remains in 
the same direction. Thus, ifa>0, then at+e>6b-+e, and 
a—c>b—e. Hence, 

A term can be transposed from one member of an in- 
equality to the other without altering the inequality, provided 
ats sign is changed. 


202. If unequals are taken from equals, the result is an 
inequality which is the reverse of the given inequality. 
Thus, ifa=y, anda> 3, then x—a<y—O. 


208. If the signs of the terms of an inequality are 
changed, the inequality is reversed. Thus, if a> 6, then 
—a<—b. (See § 33.) 


198 SCHOOL ALGEBRA. 


204, Hence, if the members of an inequality are multi- 
pled or divided by the same positive number, the inequality 
remains in the same direction, by the same negative num- 
ber, the inequality is reversed. 


(1) Simplify 4a —-3 > sz 2 
We have Agee ene 
2 5 


Multiply by 10, 402-— 30> 152—6. 
Transpose, 25.2 > 24. 
Divide by 25, a > 24. 


Therefore the value of x is greater than 24. 


(2) Find the limits of w, given 
x—-4>2— 382, 


82—-2<24+3. 
We have e—4>2-— 32a, : (1) 
and 38a—-2<2+3., (2) 
Transpose in (1), 4x >: 
Pg dae dye 
Transpose in (2), aay, 
sed ee eas 


Therefore, the value of w lies between 14 and 23. 


(3) Ifa and 6 stand for unequal and positive numbers, 
then a? + 6 > 2ab. 


Since (a — b)? is positive, whatever the values of a and b, 
(a—b)? >0. 
a?—2ab +b? >0. 
a2 +b? > 2ad. 


—— 


INEQUALITIES. 199 


Exercise 68. 


1. Simplify (v7+1)?<2#+32—65. 
2. Simplify — Fscke Pe. 


3. Simplify ++ 26 > 72. 








4. Simplify 834—2< mi 7h. 


Find the limiting values of x, given 


5. 44—6< 24+4, 
242+4>16— 22. 

2 
6. = +bn—ab > 
bx 


Lay f 


— ae + ab <=. 


Find the integral value of x, given 


8. 


7 4(@4+2)+42<4i(e —4)+3, 
4@+2)+32>3@t1) +4. 


Twice a certain integral number increased by 7 1s 


not greater than 19; and three times the number dimin- 
ished by 5 is not less than 13. Find the number. 


If the letters stand for unequal and positive numbers, 








show that 
9. a +30’? > 2b (a+ 3B). 
10. a + BF >a’) + ab’. 
1. @W4+@84+e>abtacHt be. 
12. @b+a@et+ab?+h'ce+ac’?+ be > babe. 
13. f4. 252, | 14, ees 


CHAPTER XIV. 


INVOLUTION AND EVOLUTION. 


205. The operation of raising an expression to any re- 
quired power is called Involution. 

Every case of involution is merely an example of multi- 
plication, in which the factors are equal. 


206. Index Law. If mis a positive integer, by definition 


mx OCs to m factors. 


Consequently, if m and m are both positive integers, 
(a")" = a" X a" X a” + to m factors 
=(aXa---- ton factors)(a X a +++ to n factors) 
“ois taken m times 
=axaxXa-- to mn factors. 
=a, 


This is the index law for involution. 


202 86, (0) a 
And (ab)"= ab X ab + to n factors 
=(aXa-~+ to factors)(b X b+ to factors) 


208. If the exponent of the required power is a composite 
number, the exponent may be resolved into prime factors, 
the power denoted by one of these factors found, and the 
result raised to a power denoted by a second factor of the 
exponent; and soon. Thus, the fourth power may be ob- 
tained by taking the second power of the second power; 


atk 


INVOLUTION AND EVOLUTION. BOD 


the sixth by taking the second power of the third power ; 
and so on. 7 


209. From the Law of Signs in multiplication it is evi- 
dent that all even powers of a number are positive; all odd 
powers of a number have the same sign as the number itself. 

Hence, no even power of any number can be negative ; 
and the even powers of two compound expressions which . 
have the same terms with opposite signs are identical. 


Thus, (6—a)’={—(a—)b)}?=(a-—)). 


210. Binomials. By actual multiplication we obtain, 


(a+ bP=a@+2ab+ 0’; 
(a+ be=a+3¢b+38ab’?+ 0; 
(a+ 6)*== at+ 40° + 6070’+ 406? + dt 


In these results it will be observed that: 


I. The number of terms is greater by one than the ex- 
ponent of the power to which the binomial is raised. 

II. In the first term, the exponent of a is the same as 
the exponent of the power to which the binomial is raised ; 
and it decreases by one in each succeeding term. 

III. d appears in the second term with 1 for an exponent, 
and its exponent increases by 1 in each succeeding term. 

IV. The coefficient of the first term is 1. 

V. The coefficient of the second term is the same as the 
exponent of the power to which the binomial is raised. 

VI. The coefficient of each succeeding term is found 
from the next preceding term by multiplying the coefficient 
of that term by the exponent of a, and dividing the product 
by a number greater by one than the exponent of 0. 


If } is negative, the terms in which the odd powers of 6 
occur are negative. Thus, 


202 SCHOOL ALGEBRA. 


(1) (a — bf =a — 3b + 8ab?— 8’. 

(2) (a — b)*= at — 40°) + 6070? — 4ab°+ Bt. 

By the above rules any power of a binomial of the form 
a -+- 6 may be written at once. 

Note. The double sign + is read plus or minus; and a+b means 


a+o or a—o. 


211. The same method may be employed when the terms 
of a binomial have coefficients or exponents. 


Since (a— bf =a'— 3a°b + 8ab?— B’, 
putting 52? for a, and 27° for 5, we have 
(5at— 24), 
= (52*)!— 8(52°)(2y’) + 8(52°)(2y")— (By), 
= 125 2° — 150 x*y* + 60 2°y° — 87’. 
Since (a@—b)*=at— 40° + 6a°b?— 4ab*> + Ot, 
putting 2? for a, and ty for 6, we have 
@—4y)' 
= (2°) 42 )EY) + 8) Gy) 40S Yt yy’, 
= a — Zaty + gay’ — Fey + Tey’. 


212, In hke manner, a polynomial of three or more terms 
may be raised to any power by enclosing its terms in paren- 
theses, so as to give the expression the form of a binomial. 
Thus, ) 


(1) @@+b+cP=[a+(O+oe)f, 
=a+3a(b+c¢)+38a(6+cl+(b+e)’, 
=a+8¢b+8ac+3ab’?+ babe 

. + 3a?+6?+380?e+3b7 +, 


INVOLUTION AND EVOLUTION. 203 


(2) (# —227+ 32+ 4), 
=[(0? — 20%) + (80-+4)f, 
= (a! — 2.2%) + 2 (2? — 22") (8244) + (8044), 
= 2 —42°+ 424 62t—405—16274 9274 242416, 
= #—44°+ 102*— 42° —72?4+ 244+ 16. 


Exercise 69. 


Raise to the required power : 


1. (a*)’. 11. (2? — 2), 

2. (a®d*)>, 12. (% + 8)°. 

3. ane 13. (22+1)% 
3 ab® 


14. (2m?— 1)*. 
15. (24+ 3y)°. 


4. (—5ab’c’)‘. : 
Bo (— Tay)’. 





3 a b3et\5 16. (2% — y)*. 
at Ke 
( Sy) 17. (vy — 2)". 
7. (— 2aty')’. 18. (l—2+ 2%). 
im 27.345 
ete - 19. (l— 22+ 32%) 
3 7203\4 . 
> (Greah 20. (l—a-+a’). 
10. (~+2)°. 21. (83—44+ 527)’, 
EVOLUTION. 


213. The nth root of a number is one of the m equal 
factors of that number. 

The operation of finding any required root of an expres- 
sion is called Evolution, 


204 SCHOOL ALGEBRA. 


Every case of evolution is merely an example of factor- 
wg, in which the required factors are all equal. Thus, the 
square, cube, fourth, ..... roots of an expression are found 
by taking one of its two, three, four ..... equal factors. 

The symbol which denotes that a square root is to be 
extracted is ,/; and for other roots the same symbol is 
used, but with a figure written above to indicate the root; 
thus, x/, ~/, etc., signifies the third root, fourth root, ete. 


214, Index Law. If m and m are positive integers, we 
have, (§ 206), 
(a™)” — ra hacen 
Consequently, Vinee : 
Thus, the cube root of a® is a’; the fourth root of 8la” 
is 8a*; and so on. 
This is the index law for evolution. » 


215, Also, since (ab)*= ad”, 
conversely, Vath" = ab = 0 ane 
and Vab = Vax Vo. 


Hence, to find the root of a simple expression : 


Divide the exponent of each factor by the index of the root, 
and take the product of the resulting factors. 


216. From the Law of Signs it is evident that 


I. Any even root of a positive number will have the 
double sign, +. 

II. There can be no even root of a negative number. 

For V—2* is neither +2 nor —2; since the square of 
+2=-+2", and the square of —x=-+ 2”. 

The indicated even root of a negative number is called 
an imaginary number. 

‘TI. Any odd root of a number will have the same sign 
as the number, 


INVOLUTION AND EVOLUTION. 205 





Aa 
Thus, Apia gee V— 27 min’ = — 8 mn?: 





18 yo =9y’ 
[16 2%? __ ate: 22°y* 
Sl a ek = 


217. If the root of a number expressed in figures is not 
readily detected, it may be found by resolving the number 
into its prime factors. Thus, to find the square root of 
3,415,104 : 


2°| 8415104 
426888 





11 


8,415,104 = 2° x 3? x 7x 112, 
, V8,415,104 = 22x38 x 7 X¥ 11= 1848. 


Exercise 70. 








Simplify : 
1. V4a%/. grey 21600. 1s ere 
2. Vea, 10. 7292, ie 
3. V1625y". 11. VW2438729, «18. eee 
4, V— 32a". 12. V—1728a%. pirate e020" 
5. V— 27H, 13. V— 348°. ae 
6. V25 at. 14. V8la*. 20. Af sete 
ON 15. V512a"b". ines 
8. Vb642", 16. V ain oe 216 a 


206 SCHOOL ALGEBRA. 


SQUARE Roots oF COMPOUND EXPRESSIONS. 


218. Since the square of a+0 is a’+2ab+0’, the 
square root of a?+ 2ab+ 0’ is a+. 

It is required to find a method of extracting the root 
a+6 when a’+ 2ab + 2 is given: 


Ex. The first term, a, of the root is obviously the square root of 
the first term, a*, in the expression. 
a+ 2ab+b?la+b If the a? be subtracted from the given 


a? expression, the remainder is 2ab + b?. 
2a+6| 2abd+ 8B? Therefore the second term, 0, of the root 
2ab + b? is obtained when the first term of this 


remainder is divided by 2a, that is, by 
double the part of the root already found. Also, since 


2ab + b?=(2a + b)d, 


the divisor 2s completed by adding to the trial-divisor the new term of 
the root. 


(1) Find the square root of 252?—202°y + 4 ay’. 


25 x? — 20 ay + 4aty?|5a — 2a7y 
25 x 


10a” — 2a°y|— 20a%y + 4aty? 
— 20 a%y + 4 aty? 


The expression is arranged according to the ascending powers of «. 

The square root of the first term is 5a, and 52 is placed at the 
right of the given expression, for the first term of the root. 

The second term of the root, —22%y, is obtained by dividing 
— 20a°y by 10a, and this new term of the root is also annexed to the 
divisor, 10a, to complete the divisor. 


219, The same method will apply to longer expressions, 
if care be taken to obtain the ¢rial-divisor at each stage of 
the process, by doubling the part of the root already found, 
and to obtain the complete divisor by annexing the new term 
of the root to the trial-divisor. 


INVOLUTION AND EVOLUTION. 207 


Ex. Find the square root of 
1+ 1027+ 252'+ 162° — 2425 — 2027°— 42. 
16 2° — 242° + 25 at — 202° + 10a? 424 1[408—307+ 22-1 
16 «6 
8a? — 3a?|— 2405 + 252% 
— 242°+ at 


82? — 6a? + 2x) 16 at — 2023 + 102? 
16a*—122°+ 42? 


8a°®°—62?+4e2—1]/— 822+ 6227-42741 
— 8e3 + 647—4241 


The expression is arranged according to the descending powers of z. 
It will be noticed that each successive trial-divisor may be obtained 
by taking the preceding complete divisor with its last term doubled. 





Exercise 71. 
Find the square root of 
1. 2*—82'?+ 182? —82r+1. 


2. Jat— 60° + 18? —4a-+ 4. 

3. 4a*—12a%y + 29x77? — 30xy* + 257%. 
4, 1+ 4274+ 1027+ 122° 4+ 92%. 

5. 16—96x2+ 2162? — 2162°+ 81 at. 
6. v'— 222? + 9527+ 2862+ 169. 

7. 42*—l1la2?+ 25 —122*°+ 302. 

8. Jat +49 — 122° — 282+ 462’. 

9. 49a*+1262°? + 121 — 732? — 198-2. 
10. 162*— 302 — 812’ + 242’ + 25. 

11. eee Mache alt 


208 SCHOOL ALGEBRA. 


12. 4at+40°—te4H. 


16 
4a? 40? 
LS ieee aha 
Sb ee se 
Lat abe eee eee 
Ho Gale om a a 


flee ie eee PAL 


4c, B82 41 By, yf 
16.0 eh ee 
aig hie tea 


2 
17. fan 284 a8 80-45 


18. l6a*+182°7+82'+ 47° + 4y+1. 


19, 92 82, 482% Te 49 
+ 2 4 2 as 


20. 4a?+~.—8_114 4a, 
a a 





Find to three terms the square root of 


21. a? +0. 24. 1+a. 27. 497+ 3. 
22. vty. 25. 1—2a. 28. 4—Ba4, 
23. 1+2a. 26. 47+ 20. 29. 4a?—1. 


220. Arithmetical Square Roots. In the general method 
of extracting the square root of a number expressed by 
figures, the first step is to mark off the figures in groups. 

Since 1 = 1?, 100 =10?, 10,000 =100?, and so on, it is 
evident that the square root of any number between 1 and 
100 lies between 1 and 10; the square root of any number 
between 100 and 10,000 lies between 10 and 100. In 


INVOLUTION AND EVOLUTION. 209 


other words, the square root of any number expressed by 
one or two figures is a number of one figure; the square root 
of any number expressed by ¢hree or four figures is a num- 
ber of éwo figures; and so on. 

If, therefore, an integral square number is divided into 
groups of two figures each, from the right to the left, the 
number of figures in the root will be equal to the number 
of groups of figures. The last group to the left may have 
only one figure. 


Ex. Find the square root of 3249. 


32 49 (57 In this case, a in the typical form a? + 2ab + 0? 
oe represents 5 tens, that is, 50, and 6 represents 7. 

107) 749 The 25 subtracted is really 2500, that is, a?, and the 
749 complete divisor 2a +6 is 2x 50+ 7 = 107. 


221, The same method will apply to numbers of more 
than two groups of figures by considering a in the typical 
form to represent at each step the part of the root already 
found. 

It must be observed that a represents so many tens with 
respect to the next figure of the root. 


Ex. Find the square root of 5,322,249. 
5 32 22.49 (2307 


4607) 32249 
32249 


222. If the square root of a number has decimal places, 
the number itself will have twice as many. Thus, if 0.21 is 
the square root of some number, this number will be (0.21)? 
= 0.21 x 0.21 = 0.0441; and if 0.111 be the root, the num- 
ber will be (0.111)? = 0.111 x 0.111 = 0.012321. 


210 SCHOOL ALGEBRA. 


Therefore, the number of decimal places in every square 
decimal will be even, and the number of decimal places in 
the root will be hadf as many as in the given number itself. 

Hence, if a given number contain a decimal, we divide 
it into groups of two figures each, by beginning at the 
decimal point and marking toward the left for the integral 
number, and toward the right for the decimal. We must 
be careful to have the last group on the right of the deci- 
mal point contain two figures, annexing a cipher when 
necessary. 


Ex. Find the square roots of 41.2164 and 965.9664. 


41.21 64 (6.42 9 65.96 64 (31.08 
36 9 
124) 521 61) 65 
496 61 
1282) 2564 6208) 49664 
2564 : 49664 





223. If a number contain an odd number of decimal 
places, or if any number give a remainder when as many 
figures in the root have been obtained as the given number 
has groups, then its exact square root cannot be found. We 
may, however, approximate to its exact root as near as we 
please by annexing ciphers and continuing the operation. 

The square root of a common fraction whose denominator 
is not a perfect square can be found approximately by 
reducing the fraction to a decimal and then extracting the 
root; or by reducing the fraction to an equivalent fraction 
whose denominator is a perfect square, and extracting the 
square root of both terms of the fraction. Thus, 


Rete e ee 
VP = 0.625 = 0.79057 ; 


ue N= EE 001 


Wine 


INVOLUTION AND EVOLUTION. 


211 


Ex. Find the square roots of 3 and 357.357. 


dark ied 
1 
27) 200 
189 
343) 1100 
1029 
3462) 7100 
6924 





3 57.35 70 (18.903..... 


1 


28) 257 
244 
369) 3335 

3321 





37803) 147000 


Exercise af ae 


Find the square root of 


Letteoo. 6. 
2. 1225. (ie 
3. 12544, 8. 
4. 253009. 9. 
5. 529984. 10. 


150.0625. 
118.1569. 
172.3969. 
5200.140544. 
1303.282201. 


113409 





ibe 
12. 
13. 
14. 


15. 


640.343025., 
100.240144. 
316.021729. 
454.585041. 
5127.276025. 


Find the square root to four decimal places of 


16. 10. 19,405: 
thas. 2050 (- 
18...5. 21. 0.9. 


22. 0.607. 


23. 0.521. 
24. 0.687. 


25. 
26. 


27. 


224, Cube Roots of Compound Expressions. 
of a+b is a&+38@)+ 8ab?-+ 6°, the cube root of 


Pi Bone 
$6 a9. §. 
4, 80 tae 


Since the cube 


+3076 + 38a?+ 6 is at. 


It is required to devise a method for extracting the cube 
root a+6 when a’+3a’b+ 3ab’-+ 6 is given: 


212 SCHOOL ALGEBRA. 


(1) Find the cube root of a?+ 3a’) + 3ab’ + 8°. 


a+ 3a7b +3ab?+ Bla+b 
Bae os as 


+3ab +6? 3.0a7b + 3ab? + 68 
3a?+ 3ab + 6? 3a7b + 3ab? + 6 


The first term a of the root is obviously the cube root of the first 
term a’ of the given expression. 

If a® be subtracted, the remainder is 3 a?) + 3ab? + 6°; therefore, 
the second term 8 of the root is obtained by dividing the first term 
of this remainder by three times the square of a. 

Also, since 307) + 3ab? + B= (3a? + 3ab+0?)b, the complete 
divisor is obtained by adding 3ab + 0? to the trial-dimsor 3a’. 


(2) Find the cube root of 82°+ 362°y + 54 ay’ + 2777. 


8a? + 36 x7y + S4ay? + 27y?|2e + 3y 
122? 8 x3 


(62+3y)3y= 182y +94? 36 ay + 54 ay? + 27 y 
12a?4+18ay+9y?|_ 36 a%y + 54 ay? + 27° 
The cube root of the first term is 22, and this is therefore the first 
term of the root. 
The second term of the root, 3 y, is obtained by dividing 36 a?y by 


3 (2a)? = 1247, which corresponds to 3a? in the typical form, and is 
completed by annexing to 122? the expression 


{3(2x)+3y33y = 18ay + 9y?, 
which corresponds to 3.ab + 6? in the typical form. 


225. The same method may be applied to longer expres- 
sions by considering a in the typical form 3a’?+ 3ab+ 6? 
to represent at each stage of the process the part of the root 
already found. Thus, if the part of the root already found 
is «+ y, then 3a’ of the typical form will be represented 
by 3(2+ y)?; and if the third term of the root be +z, the 
3ab-+ &? will be represented by 3(a+y)z+2%. So that 
the complete divisor, 3a’?-+ 3ab + 0’, will be represented by 


d(a+tyyl+3(e#+y)z242. 


INVOLUTION AND EVOLUTION. 213 


Find the cube root of 2°— 32°+ 52°— 3a —1. 








x2? — 7+. I 
°° — 32° + 52° — 32-1 
3 at x 
(3a? — x)(—2) = soe + —32° +525 - 
3a%*—S8a5+ a71/—325 +3at— 2 





—3at+ 623-32 —-1 
3 (a? — x)? =3at— 623 + 3.2? 
(3 2?—3a—1)(—l)= —3a? +3a¢ +1 


ed Ll ok + bao a 





The root is placed above the given expression for convenience of 
arrangement. 

The first term of the root, x?, is obtained by taking the cube root 
of the first term of the given expression; and the first trial-divisor, 
3x4, is obtained by taking three times the square of this term. 

The first complete divisor is found by annexing to the trial-divisor 
(3a? —a)(—«), which expression corresponds to (3a+0)6 in the 
typical form. 

The part of the root already found (a) is now represented by 2?—«; 
therefore 3a? is represented by 3(a?— 2)? =3at—623 + 327, the second 
trial-divisor; and (3a + 6)b by (83a?—3a—1)(—1); therefore, in the 
second complete divisor, 3a? + (3a + b)bd is represented by 


(3 a*—6 a3 + 3a”) + (—3a?—32—1)x (—1)= 3at— 603 + 3241. 


Exercise 73. 
Find the cube root of 
a+ 8a + 8a27+ 2%. 
8+122+62'+ 2. 
x’ — Baz’ + 5a®z*®— 8a°x — af. 
1— 62+ 212?— 442° + 682*— 542° + 27 2x® 
1— 382+ 62? — 72°+ 62*— 32°+ 2° 
2t+1— 62—62°+ 1527+ b2*— 202%. 


oO nT Pw YP 


214 SCHOOL ALGEBRA. 


7. 642°—1442°+ 8— 8642+ 1022?— 171 2°+ 204 2%. 
8. 27a°— 27a°— 18a'+ 17a°+ 6¢’— 8a—1. 

9. 82°— 362°+ 662‘ — 682° + 8382°—9x+1. 

10. 27+ 1082+ 902?— 802? — 60 2* + 482°— 82° 


. i dietel A 
Lie ol a 
pitas een 
Toe OL AOL ee 
ae a oe? 


226. Arithmetical Cube Roots. In extracting the cube root 
of a number expressed by figures, the first step is to mark 
it off into periods. 

Since 1 =1?, 1000=10*, 1,000,000 = 100%, and so on, it 
follows that the cube root of any number between 1 and 
1000, that is, of any number which has one, two, or three 
figures, is a number of one figure; and that the cube root 
of any number between 1000 and 1,000,000, that is, of any 
number which has fowr, five, or siz figures, is a number of 
two figures ; and so on. 

If, therefore, an integral cube number be divided into 
groups of three figures each, from right to left, the number 
of figures in the root will be equal to the number of groups. 
The last group to the left may consist of one, two, or three 
figures, 


227. If the cube root of a number have decimal places, 
the number itself will have three tumes as many. Thus, if 
0.11 be the cube root of a number, the number is 0.11 x 0.11 
x 0.11 = 0.001331. Hence, if a given number contain a 
decimal, we divide the figures of the number into groups 
of three figures each, by beginning at the decimal point 
and marking toward the left for the integral number, and 


INVOLUTION AND EVOLUTION. 215 


toward the right for the decimal. We must be careful to 
have the last group on the right of the decimal point con- 
tain three figures, annexing ciphers when necessary. 


228, Notice that if a denotes the first term, and 6 the 
second term of the root, the first complete divisor is 
3807+ 8ab-+ b’, 
and the second trial-dwisor is 3(a-+ 6)’, that is, 
80+ 6ab +36, 
which may be obtained by adding to the preceding complete 
divisor zts second term and twice rts third term. 
Ex. Extract the cube root of 5 to five places of decimals. 
5.000 (1.70997 | 


1 
3 x 10? = 300 4000 
3(10 x 7) = 210 
= 49 
559 3913 
259 87000000 







3 x 1700? = 8670000 
3(1700x9)= 45900 


OF es 81 
8715981 


45981 
3 X 1709? = 8762043 








78443829 

85561710 
78858387 
67033230 
61334301 










After the first two figures of the root are found, the next trial- 
divisor is obtained by bringing down the sum of the 210 and 49 
obtained in completing the preceding divisor; then adding the three 
lines connected by the brace, and annexing two ciphers to the result. 

The last two figures of the root are found by division. The rule 
in such cases is, that two less than the number of figures already 
obtained may be found without error by division, the divisor being 
three times the square of the part of the root already found. 


216 SCHOOL ALGEBRA. 


Exercise 74. 
Find the cube root of 
1. 4913. 3. 1404928. 5. 3885828.352. 
2. 42875. 4, 127263527. 6. 1838.265625. 
Find to four decimal places the cube root of 
(Peeve 9. 3.02. 11. 0.05. 13. 2. 15. 3%. 
Biol) 10. 2.05. 12. 0.677. 14. 3. 16. +s. 


229, Since the fourth power.is the square of the square, 
and the sixth power the square of the cube, the fourth root 
is the square root of the square root, and the sixth root is 
the cube root of the square root. In like manner, the 
eighth, ninth, twelfth, ...... roots may be found. 


~ 


Exercise 75. 
Find the fourth root of 
1. 8lat+ 1082? + 542*°+ 122-41. 
2. 162* — 82a2° + 24077? — 8a*x + at. 
~ 1+4e7+4+42'+ 1028+ 162? + 102?+ 192*+ 162". 


J%) 


Find the sixth root of 
4. 1+6d+d°+6d*°+15d‘*+ 20d*+ 15d’. 
. 729—14582+12152? —540234 135 2*—184°+ 2° 
1—18y+ 1357 — 5407? + 12157‘ — 14587 + 729%. 


oO oa 


CHE Pighorn Ve 
THEORY OF EXPONENTS. 


230. If nm is a positive integer, we have defined a” to 
mean the product obtained by taking a as a factor m times. 
Thus a* stands for axaxa; 6* stands for bX bx6x 6. 


231, From this definition we have obtained the following 
laws for positive and integral exponents : 
Dad Xiet=aaet™ 
IL (a™)"=a™, 
III. us Bee anid 97 TUT. 
a 
LV Van = a™. 
Wie LOLs sa any™: 


232. Since by the definition of a” the exponent n denotes 
simply repetitions of a as a factor, such expressions as a3 and 
a~* have no meaning whatever. It is found convenient, 
however, to extend the meaning of a” so as to include 
fractional and negative values of n. 


233, If we do not define the meaning of a" when 7 is 
a fraction or negative, but require that the meaning of a” 
must in all cases be such that the fundamental index law 
shall always hold true, namely, 

q™ x q” = Cmte: 
we shall find that this condition alone will be sufficient to 
define the meaning of a” for all cases. 


218 SCHOOL ALGEBRA. 


234. To find the Meaning of a Fractional Exponent. 


Assuming the index law to hold true for fractional expo- 
nents, we have 


3 3 3 3 iz 
at x at x xataatth lt gi =e 


4 : = he to n terms ms 

A X Ah wee to n factors = an'2™™” = 4 en 
x = os < = ton terms ter 

Gn X QA vee to n factors =an ‘2 == 8 sae 


That is, a? is one of the two equal factors of a, 
a? is one of the three equal factors of a, 


kes 

a‘ is one of the four equal factors of a’, 
1 : 

an is one of the m equal factors of a, 


an is one of the n equal factors of a”. 
Hence at= Va; at = Va; 
at= Va; a® am (§ 218) 


1 1 1 
Also, am X ar X an Xe to m factors. 


1 i 1 m 
—~+-—+-—.... to m terms -- 
=—— (im ey ——i(]ien 


“aa (Va). 


The meaning, therefore, of an. where m and are posi- 
tive integers, is, the nth root of the mth power of a, or the 
mth power of the th root of a. 

Hence the numerator of a fractional exponent indicates 
a power, and the denominator a root; and the result is the 
same when we first extract the root and raise this root to 
the required power, as when we first find the power and 
extract the required root of this power, 


THEORY OF EXPONENTS. 219 


235. To find the Meaning of a’. 
By the index law, 


me eer ees OL: 


*. a’ =1, whatever the value of a is. 


236. To find the Meaning of a Negative Exponent. 


If n stands for a positive integer, or a positive fraction, 
we have by the index law, 
OG AY ipa Eee god 
But ural, 
.axar=i. 


That is, a” and a~ are reciprocals of each other (§ 167), 
so that a” = as andia == as 
a a 
237. Hence, we can change any factor from the numerator 
of a fraction to the denominator, or from the denominator 


to the numerator, provided we change the sign of its exponent. 


Bas La 
abd? 





Thus ae may be written al’c*d~™, or 

238. We have now assigned definite meanings to frac- 
tional and negative exponents, meanings obtained by 
subjecting them to the fundamental index law of positive 
integral exponents; and we will now show that Law II., 
namely, (a”)”= a", which has been established for positive 
integral exponents, holds true for fractional and negative 
exponents. 


(1) Ifn is a positive integer, whatever the value of m, 


We have (a™)n = a™ xX a™ x a™..... to n factors, 


= gm+m+m eevee to n terms 
J 


= amr, 


220 SCHOOL ALGEBRA. 


(2) If nis a positive fraction 2, where p and q are posi- 
tive integers, we have q 


(amr = (amy = Vamp 2 234 
= Vam (1) 
exe ae % 234 
= q™ xa 
=a, 


(3) If is a negative integer, and equal to —p, we have 


1 
(a™)" = (a™)- P= (amp 2 236 
1 
<a (1) 
=a-™ 2 236 
= q™(—Pp) 
=qmn, 


(4) If nis negative and equal to the fraction ae where 
p and qg are positive integers, we have d 
1 











Pp 
(am)* = (a")g= ——— 3 236 
(a™) 9 
ee ae @ 234 
a™ 
a + @ 234 
ad 
Pogtl 
q™ x= 
reel 
i qm(—n) 
1 
a qQg mn (1) 
= gmn, 2 236 


Hence, (a”)"=a™", for all values of mand n. 


THEORY OF EXPONENTS. 221 


239, In like manner it may be shown that all the index 
laws of positive integral exponents apply also to fractional, 
and negative, exponents. We will now give some exarples. 


(6) @tyaat Da g2 = 
Qa 


(6) sie ee a he 
a 


(1) Vabtcotd= akb-1¢- 3 3, 











! 


(9) (16a) *_ (81a \t _ 27 bt _ 27 athe 
818 16a 











2 _2 
0 GO Serine ake 7 aan 


Exercise 76. 
Express with fractional exponents: 
De Var © 3. Vale 8. Va. 7. VatV24+V 1684 
pee ed, >) 8. <6. Va’. 8.9 Vax? + Vato. 





222, SCHOOL ALGEBRA. 


Express with root-signs : 











3 os hey Fee fae Pac 
9.5.04. 11. a*b8; 13.007. 15. a?— 2% ¢8. 
2 1,2 2 8 2 2 2 
10. 3, 12. a5bd8, 14. 32%y 4. 16. aS + x55, 
Express with positive exponents: 
—1 ,.2 
17. a-*. 19. Bay’. 21. 4aty, 23, ee 
By gh 
sisya 5, . 20. 4 ye ae Oe 3a-*b?. 
Write without denominators: 
2 
2,3 a 
yy eka a6. eee, 28.4 
a eed aren | rae Ais. 
2 —1 },-2 ,-3 —4f—-5 ,—6 
ST ees. 27. een 29. OVE 
SY CEUs C7 Open 
Find the value of 
5 2 3 2 5 
SiR ema y peyE 34, 36, 36. (—27)5 x 257, 


31. 16-2. 33. (—8)-#. 35. (—27)8. 37. 817? x 164. 


Simplify : 
38. 8? x 472, 40. Gis x (ay)? 42. (a $B)" : 
39. (se)? x 1674. 41. wa 1bF x atdt. 48. (a 25) 


If a=4, b=2, c=, find the value of 
44. ab 46. a 20%, 48. 8(ab)*. +50. (ad*e)t. 


45. ab. 47. ate 8. 49. 2(ab) 4%. 51. (ab%)?, 


THEORY OF EXPONENTS. 223 
240. Compound expressions are multiplied and divided as 
follows : 


1 


: 1 Ad 1 aL ae 1 
(lye Multiply 2? + 2*7* + 4% by 2? —2x77* +77. 
1 Tot a 
Me ak oy i) 
1 1 1 1 
24 — pFy* + 74 
he HI 1 Us 
ary) Pet ys = 44/2 
Sip Ug 
+ aby? + atyd + y 


x + why? +y 
(2) Divide V2?+ Vx—12 by Vz—8. 
aot 4 at —12|at — 3 





ai — 348 at +4 
+4a3 — 12 
+4a% — 12 








Exercise 77. 


Multiply : 
1. at +53 by at — 32. 3. at —b? by a? —}?. 
2. at +b? by at + 82. 4. ee by xt — 2m. 


Bp aty ty? by ot aty ty 
, ee 
a tay +y* by at —y. 
eee po b-* by 1 —b 248-7. 
ee Ady by at yt +42: 
10. «86-24 2at — 383 by 20-4? 4a°# 6a 288. 


3 


224 SCHOOL ALGEBRA. 


Divide : 
11. a—b by at — BF. 13. a—b by a* — BF, 
12. a+6 by at + 8. 14. a+b by ak +55. 


15. Qa-8+ 62 y — 162°y— by Qa4+2a%y 4 4ahy 

16. ety tz—S8atyte by bt yo. 

17. 2-323 +4+82'—1 by a1. 

18. statytty by a — at ot + of? 

19. 2 —4e5+146a 3 by 22. 

20. 92—122? 2444 2427 by 8222-273. 
Find the square root of 

21. x 42341, 23; x — Ags + 4, 

22. 4at—4a3bt+ 9°. 24. 4a7+4a7+1, 

25. 9a—12a?+10—4a F447, 

26. 49a? — 282+ 1823 —423+1. 

27. m+ 2m—1—2m "+m. 

28. 144y7'—2y-8— 4474 253 — 24 $4 164, 
Expand : 

29. (w@—a). 31. (2at—a-®). 33. (LV 2 —4-V2)'. 

30. (Vai—42). 32. (Qa2%+a%). 34. (AVa%+427) 


CHAPTER XVI. 


RADICAL EXPRESSIONS. 


241, A radical expression is an expression affected with 


the radical sign; as, Va, V9, Va, Va +6, V32. 


242, An indicated root that cannot be exactly obtained 
is called a surd, or irrational number. An indicated root that 
can be exactly obtained is said to have the form of a surd. 

The required root shows the order of a surd; and surds 
are named quadratic, cubic, biquadratic, according as the 
second, third, or fourth roots are required. 

The product of a rational factor and a surd factor is 
called a mixed gurd; as, 83V2, bVa. The rational factor 
of a mixed surd is called the coefficient of the radical. 

When there is no rational factor outside of the radical 
sign, that is, when the coefficient is 1, the surd is said to be 


entire; as, V3 2, Va. 


243, A surd is in its simplest form when the expression 
under the radical sign is integral and as small as possible. 

Surds which, when reduced to the simplest form, have 
the same surd factor, are said to be similar. 


Norr. In operations with surds, arithmetical numbers contained 
in the surds should be expressed in their prime factors. 


REDUCTION OF RADICALS. 


244, To reduce a radical is to change its form without 
changing its value. 


226 SCHOOL ALGEBRA. 


Case I. 


245. When the radical is a perfect power and has for an 
exponent a factor of the index of the root, 


(1) Vai=at=at= Va; 
(2) V36 0%? = V(6ab)? = (6ab)? = (6.ab)? = V6ab; 
(3) V25 abd = V(b abc! = (5 arbct)§ = (5 arbe)s 
= V5 abet. 
We have, therefore, the following rule: 
Divide the index of the root by the exponent of the power. 


Exercise 78. 








Simplify : 
ia \/ 25. 6. V/ 02d. We Nee 
2. VTC. ses Wri ee 
6/57 8/474 13:., 4) Loge 
Shan 27! 8. Va'dbt. (x — 3) 
4. \/49, 9. \/ 27 a8b®. ey xe) 
ee 13. 6} @0° 
5. V64. 10. V16a‘d'. 8 xy? 
CasE II. 


246. When the radical is the product of two factors, one of 
which is a perfect power of the same degree as the radical. 


Since Va"b = Va" x Vb =aVb (§ 215), we have 
(Oiney Woe Xa ee ny ae 
(2) W108 = -V27 x 4 == V27 x V4 =38V4; 


RADICAL EXPRESSIONS. DOT 





(3) 4V 72020? = 4-36.00? x 26 = 4-V36 ab? x V26 
=4x 6abV2b = 24ab V26- 

(4) 2V54 ab = 2V27 a? x Zab = 2V 2708 x V2ub 
= 2x 8aV2ab = 6a V2ab. 


We have, therefore, the following rule: 


Resolve the radical into two factors, one of which vs the 
greatest perfect power of the same degree as the radical. 

Remove this factor from under the radical sign, extract 
the required root, and multiply the coefficient by the root 
obtained. 


Exercise 79. 














Simplify : 
1. V28. 13. 7V144. os. 3| O425y 
Oye) F 3,3 
Sioa 12. 14. 8Vmn. 27min 
3, -V72. 15. 3VBb%. 26. te 
4. ~/500. 16. 2Wakc*, a 
ee ov. 4: 125 2° 
5. W482. t2 L1lVa®, : 216y3 
6. 192. 18. 7V8a°d. 
ae BT ap. 24 
7. W128. 19. 6V27 mn. 943 
8. 243. 20. 4Va"y’. oo tl aut 
apes — V 1296 
9. V176. 21. 1029. 
Lae 
10. 405. 22. /— 2187. 30. yee. 
Tre —— 
11. BV 112. 23. 1250. 3ab_ [B02 


12. 3-864. 24, 4r/648. ole OWN Gait 


228 SCHOOL ALGEBRA. 


Case III. 


247. When the radical expression is a fraction, the denomi- 


nator of which is not a perfect power of the same degree as the 
radical, 


5 10 eT VOX 

<—=~4/--=4/10 ae 

\ 7 Sr 10. 

Ge Tee we 9150 
tess ee —=—sx = * Ty 
Ne aie 4x9 * 36 eval; 


seo aeest et 35x 8x4 : 1 | 
2 Se ae 
a> a= 27 x8 27 x 8 





We have, therefore, the following rule: 


Multiply both terms of the fraction by such a number as 
will make the denominator a perfect power of the same 
degree as the radical; and then proceed as in Case LT. 


Exercise 80. 











Simplify : 
ues a Mein # Tan as 10. 2V3%. 
yh se. 5B. W238. 8. V2. Lea 
Beet £ 6.08 Pr. SimeNate, 12. ee 
ate sf cat acy? 
1Sii | se 15. 17., Ales 
5 NG Ne 
14 Abe 16. 412. 1g 
a® ~ N125¢ 3x’ y2* 


RADICAL EXPRESSIONS. 929 


Case IV. 
248. To reduce a mixed surd to an entire surd. 
Since aVb = Va" x Vb = Va"b, we have 
(1) 8V5=V3'x5=V9x5=V45; 
(2) vbVbe = V(b) x be = Vail? x be = Vaibic : 
(3) 2aVay = V(2a) Xx cy = V8e xX ay = VB ay ; 
(4) 8yVei=VBy)! xX @ = VB1y 0". 


We have, therefore, the following rule: 





Fase the coefficient to a power of the same degree as the 
radical, multiply this power by the given surd factor, and 
dicate the required root of the product. 


Exercise 81. 


Express as entire surds: 
1. 5V5. 5. 2N/8. CD ean aia kame SVE 
evil 6.81/72.  10..— 8V7. 14. —3-Val. 
3. 33. Fen eel l nh / 1004 15... Bach 
4, 2V/4, Base 12.0 D4) 2. 16. —3V mi. 


Case V. 
249. To reduce radicals to a common index. 


(1) Reduce V2 and V3 to a common index. 
V2 = 2 = = V8 = V8. 


43 = 3t 38 YR = V5. 
Hence, 


230 SCHOOL ALGEBRA. 


Write the radicals with fractional exponents, and change 
these fractional exponents to equivalent exponents hawng the 
least common denominator. Raise each radical to the power 
denoted by the numerator, and indicate the root denoted by 
the common denominator. 


Exercise 82. 


Reduce to surds of the same order : 


1. V3 and V5. 7. V2, V8, and V5. 

2. V14 and V6. 8. Va?, Vb, and Ve. 

3. -/2and V4. 9. Vai, Ve, and V2". 

4. Vaand Vb". 10. V2y, Vabe, and W2z. 
5. V5 and V7. ll. V2—yand Va+y. 
6. 22 23 and 23. 12. Va+b and Va—b. 


Nore. Surds of different orders may be reduced to surds of the 
same order and then compared in respect to magnitude. 


Arrange in order of magnitude : 


13. V15 and V6. 15. /80, V9, and V8. 
14. V4 and V3. 16. V3, V5, and V7. 


ADDITION AND SUBTRACTION OF RADICALS. 


250. In the addition of surds, each surd must be reduced 
to its simplest form; and, if the resulting surds are similar, 


Find the algebraic sum of the coefficients, and to this sum 
annex the common surd factor. 

If the resulting surds are not similar, 

Connect them with their proper signs. 


RADICAL EXPRESSIONS. ak 


(1) Simplify V27 + -V48 + V147. 
V27 = (3? x 3)2=3 x 33 =3V3; 
V48 = (2! x 3)8 = 2? x 32 = 4 x 38 = 43; 
V147 = (7? x 8)? =7 x 32 = 7V3. 
o. V27 4+ V48 + V147 = (8444 7)V3 =14V3. Ans. 
(2) Simplify 2/320 —8-~/40. 
2320 = 2(2 x 5F=2xK Bx 5E=8V5; 
3V40 = 3(2 x 5)F=3 x 2x 58 = 6 V5. 
“. 2320 — 340 =(8 —6)V5=2V5. Ans. 


(3) Simplify 2V3—8V34 V+. 








2V5 = 2V/18 = 2V15 xi =2Vv 15; 
3V8 =3V15 =3V15 x ge = 3 V15 

z= ./4x 15 \ 4 

. Se cs se 
Vi = 1p 15 Xx ip = 1s V15 


1 2V8 —3-V8 4+ VA = (8-8 + 3%) V15 =i VI15. Ans. 


Exercise 83. 

Simplify : 

Po 4 bas y 116/11: 

MeN) 8 = 54/8 95/8. 
pny Ae PX/89— ~/108. 1. V8 43/4844 75. 
. BV2+44V72— V/64. 8. 4V147+ 3775+ -V192. 
.1AW54+21V5414V40. 9. Vativat va. 
; 8VB— 548+ -V248. 10. Va?+ iWVai—8V 27a". 


Zor SCHOOL ALGEBRA. 


Lis Val OW Ge Ow 

12. V25b+ 2V96b — 8V46. 

13. 2/175 — 8-V638 + 5-V28. 

14, V2+ 8V382-+ 1/128 — 6V/18. 

15. V75 + V48—-V 147 + -V800. 

16. 20-V245 — V5 + -V125 — 24-180, 

17. 2V20-+ 4VI13 — 2-V87 +. SV — VIB. 
18. 725+ 4745 — V9 — 2-804 V20—4 V64, 
19. V54+ VE— V250 — 8-v32. 

20. 2V8+ V60— V15+ V3+ V4. 

a1. VOT — V8 + V125e. 

22. Vabb — Vb'+ W320. 

23. Vata + Vor —-V4a°b'x. 

24. V4a%pe t+ Vy't2+ Va'x. 

25. Vale —av4e+bvare. 

26. V8la’— V16a+ V256a°. 

27. W27m*t — V125m + V216 m. 

28. V8a—V50a?— 8V18a. 

29. 6aV63ab) — 8-V1120°8? + 2abV348 ab. 
30. 3V125 min? + nV 20m — -V500 min’. 
31. V32a'b® + 6V726 + 8-V128 ab? 

32. 2N/a% — 8 0V646 + bavVatb + 2aV125 6. 


22. 
23. 
24. 
25. 


RADICAL EXPRESSIONS. 


233 


MULTIPLICATION OF RADICALS. 


251, Since Va x Vb = Vab, we have 
fms x oV2—3 X 5 xXV/8.x-V2 = 15V16= 60; 
(2) 83V2 x 4V3 =38V8 x 4W9 = 12V72. 

We have, therefore, the following rule: 


Express the radicals with a common index. Find the 
product of the coefficients for the required coefficient, and the 
product of the surd factors for the required surd factor. 


Reduce the result to its sumplest form. 


. V8x V97. 
. VEX -V20. 
oV2x N18. 
. VBxXV9. 
BN \/ 52: 
ETE aN ES 


Beaise oh 
7. VW4x V8. 
8. V27x V9. 
9. V2x VI12. 
LO ry avo, 
11. V3xVI18. 
12. V6x V8. 


13 


. Wd4 x V9. 
2 2/8x V2: 
. V8xV-=4, 


Vix V¥—49. 
. V8lxV—45. 


. 2V18 x 8V3. 


19. (V18 + 2V72—8V38) x V2. 
20. (W732 —1V8644 8V4) x V2. 
a. (AV 27 — 1/2187 + 4-482) x V3. 


V5 x V4. 


V64 x V16. 
V3 x V7. 


V16 x 0/250. 


26. V2 x Vi. 
OAD gt GN) 
28. V81x V3. 


29 GaN ROGAN) Fs 


30 


31 


32 


33 


VEX VE. 
. V2a x V2. 


Vy XN oY. 
VIX V5. 


234 SCHOOL ALGEBRA. 
252. Compound radicals are multiplied as follows: — 


Ex. Multiply 2V34+8V2 by 8V38—4Vz. 
2V3 + 3V x 
3V3 —4vV 2 
18 +9V32 
—~8V3e—122 
18+. V32—122 


Exercise 85. 


Multiply : 
1.V5+V4by V5—V4 4. 84+8-V2 by 2— V2. 
2. V9—VI7by V9+VI17. 5. 54+2V3 by 8—5V3. 
32 3-2 bby Be o: 6. 8— V6 by 6— 8V6. 
7. 2V6 —8V5 by V3 + 2V2. 
Bie Tey ODN ee 
9. V9 — 2/4 by 4734+ V2. 

10. 2V30—8V5+ 5V38 by V8+ V3 — V5. 

11. 8V5 — 2V38 4+ 4y/7 by 83V7— 4V5 = Dae 

12. 4V8+1V12 —1V32 by 8-V32 —4V50 — 2v2. 

13. V6 —-~V/3 +4 V/16 by 36+ V9 — V4, 

14, 2V2—8V84 8-V§ by 8V3— V12 — V6. 

15. 2V3 —4V8 — 7V8 by 8-V3 — 5-V30 — 2-48. 

16. 2V124 8V34 6V32 by 2V12 + 8-V3 4 6V1. 


RADICAL EXPRESSIONS. 200 


DIvIsIon oF RADICALS. 
258, Since SO = Vb, we have 
Va Va 


4/8 _ 
O) OV 








4/3 40/3? 4/3? x DB 8/75, 


=. i 


JN BION iy aa 
We have, therefore, the following rule: 





(2) 


Express the radicals with a common index. Find the 
quotient of the coefficients for the required coefficient, and the 
quotient of the surd factors for the required surd factor. 

feduce the result to rts sumplest form. 


Exercise 86. 


Divide: 
Dey ote by V3. 4. Vi by Vz. 7. V4¢ by V2. 
2. V81 by V3. 5. VE by V3. 8. WBF by V5E. 
pm apuby Va’... 6. Vy, by Ve. 9. V$2 by VEE 
10. 83V6+ 45-V2 by 8v3. 
11. 42/5 — 30V3 by 2V15. 
12. 8415+ 168V6 by 3V21. 
13. 3074 — 36-V10+ 30-V90 by 3/20. 
14, 50V18+ 18-V20—48V5 by 230. 
15. V54 by V36. 17. VI2by V6. 19. V2 by W338. 
16. V49 by V7. 18. V& by V6. 20. V2z by Vz. 
21. +/0.064 by V/10. 22. Vat—y* by x+y. 


236 SCHOOL ALGEBRA. 


254. The quotient of one surd by another may be found 
by rationahzng the divisor; that is, by multiplying the 
dividend and divisor by a factor which will free the divisor 
of surds. 


255. This method is of great utility when we wish to 
find the approximate numerical value of the quotient of 
two simple surds, and is the method required when the 
divisor is a compound surd. 

(1) Divide 8V8 by V6. 

8V8_6V2_6V2xVv6_ 6V12 


Ve Ve Teves r = V12 = 2V3. 


(2) Divide 8V5—4V2 by 2V548V2. 


8V5—4V2_ (3V5—4V2)(2V5—3V2)_54—-17V10 
2V54+3V2 (2V543V2)(2V5—-3V2) 20-18 


—4=17V10 _ 97 _ gi V0, 





wt 2 
= 5 
(3) Given V2 = 1.41421, find the value of Rie 
BSA RPE A) Scilla 
V2 V3xv2 2 2 


Tacidar Exercise 87. 


1. Va+vob by Vab. 7. 3+5V7 by 8—5V7. 
Bee 195 bys 57 6b: 8. 21V3 by 4V8—8v2. 
8. 8 by 11+3V7. 9. 75V14 by 8V2+4 2V7. 
4. 8V2—1 by 8V2+1. 10. V5—-V38 by V5 +-V3. 
6. 17 by 8V7+2-V8... 115 V8 + V7 by eee 
6. 1 by V2+ V3, 12, 7—3V10 by 5+4¥V5. 


RADICAL EXPRESSIONS. 937 


-Given V2=1.41421, V8 =1.73205, V5 =2.23607; 
find to four places of decimals the value of 


13. 


14. 


15. 





eLO «16. fab 19. mite 22. {crest wes 
V2 V500 8/2 5+ 4/5 
Pen a) = og, BE VS. 
V3 V/ 2438 V/125 x/5 73 
12 18. 1 = 21. ahs 24. BV2—1 mak 
V5 23 4/5 3V2+41 bel 


INVOLUTION AND EVOLUTION OF RADICALS. 


256. Any power or root of a radical is easily found by 


using fractional exponents. 


i. 


2. 


(1) Find the square of 2Va. 
(2V a) = (2a8)? = 2a? = 408 = 4V a. 
(2) Find the cube of 2-Va. 
(2Va)3 = (2.02) = 2a? = 8a? = 8ava. 
(3) Find the square root of 4a-V ad’. 
(4. -V 0363)? = (4. 7a2b2)3 = 42 ett? = 43 atadd = OV aOR, 


(4) Find the cube root of 4a-Va'd'. 
(4 @-Va3b3)3 = (4 wa2h?)8 = 43 aha 3ht = 46 aba8h8 — V/16 abe2?. 





Exercise 88. 


Perform the operations indicated : 
ao oe rt Pepanene eaci 
(Vm). 8. (Wat). 5. VV ye 
(Vm 4. (Vy, 6. Va 5*. 


238 SCHOOL ALGEBRA. 





Teh Bebo yien 10. Aas 13. \/Ga— 2b)", 
fb DE br 2 4 ee af. aoe 

8. (Vaer—y) 11. VV729. 14. V*/320%. 

9. (Wa). 12. \V158. 15. V128-/ 248 a", 


PROPERTIES OF QUADRATIC SURDS. 


257. The product or quotient of two dissimilar quadratic 
surds will be a quadratic surd. Thus, 


Vab x Vabe = abvVe; 
Vabe+ Vab = Ve. 

For every quadratic surd, when simplified, will have 
under the radical sign one or more factors raised only to 
the first power; and two surds which are dissvmilar cannot 
have adl these factors alike. 

Hence, their product or quotient will have at least one 


factor raised only to the first power, and will therefore be 
a surd. 


258. The sum or difference of two dissimilar quadratic 
surds cannot be a rational number, nor can tt be expressed 
as a single surd. 

For if Va+ V6 could equal a rational number c, we 
should have, by squaring, 


at2Vab+b=e'; 
that is, +2\/ db Or Oe. 
Now, as the right side of this equation is rational, the 
left side would be rational; but, by § 257, Vab cannot be 
rational. Therefore, Va+ Vé cannot be rational. 


In like manner, it may be shown that Va + Vé cannot 
be expressed as a single surd Ve. 


RADICAL EXPRESSIONS. 939 


259, A quadratic surd cannot equal the sum of a rational 
number and a surd. 


For if Va could equal ¢-+ Vb, we should have, by 
squaring, 
a=e+2cVWb+, 
and, by transposing, 
9eVb =a—b—e*. 
That is, a surd equal to a rational number, which is 
impossible. 


260. Tf at+Vb=2+Vy, then a will equal x, and b 
will equal y. 

For, by transposing, Vb —Vy=2—a; and if 5 were 
not equal to y, the difference of two unequal surds would 
be rational, which by § 258 is impossible. 

pe a) enor 2. 

In like manner, if a— Vo=x2— Vy, a will equal z, 

and 6 will equal y. 


261. To extract the square root of a binomial surd, 
Ex. Extract the square root of a+ Vo. 


Suppose Va+Vb=Vi+ Vy. (1) 
By squaring, a+ Vbsa4+2Vay +y. (2) 
.a=a+y and Vb =2V2y. 3 260 
Therefore, a—Vb=x-—2Viy+y, (3) 
and Va—vVb=vVz— Vy. (4) 
Multiplying (1) by (4), 
V@e—b=2-y. 


But a=x2+y. 


240 SCHOOL ALGEBRA. 


Adding, and dividing by 2, «= 2+ V¢=% vat 


Subtracting, and dividing by 2, 


a—vVa?—6b 


= 


; ; 
: an Vetveas a|a— Va? —8, 
“ Va+vo POL me 5 


From these two values of xz and y, it is evident that this 
method is practicable only when Bee b is a perfect square. 


(1) Extract the square root of 7+4-V3. 


Let Va + Vy =V7+4V3. 
Then Vi —-Vy =V7—4v3. 
Multiplying, e—y = V49 — 48. 
“ £—y= 1. 
But et+y=7. 
*. w©=4, and y=3. 


o Ver Vy =24 v3. 
0 V744V8=24 V3. 


A root may often be obtained by inspection. For this purpose, 
write the given expression in the form a + 2V, and determine what 
two numbers have their sum equal to a, and their product equal to 0. 


(2) Find by inspection the square root of 75 —12-V21. 


It is necessary that the coefficient of the surd be 2; therefore, 
75 —12V21 must be put in the form 
75 —2V 756. 


The two numbers whose sum is 75 and whose product is 756 are 
63 and 12. 


Then 15 — 2756 = 63 + 12 — 2V63 x 12, 
= (vV63 ~— V12)%. 
That is, V63 — V12 = square root of 75 —12V21; 


or, 3V7 — 2V3 = square root of 75—12V21. 


RADICAL EXPRESSIONS. 941 


Exercise 89. 


Find the square root of 


ey eae Bee 7. 16-2 5v7. 13. 94-4. 49/5. 
See ee / 72. Be Yo on/ et 14. 1T — oN /80 


3. 7-42-10. IS 8-7/3: 15. 4744/38. 
4. 18+ 8V5. LOSS Sy/6'-— 90. 16.45.29 2)- 60/22, 
5. 8412/15. rH Gass a Lotaw © ems Wet oe ee EET, 
Geto 4/14, 12. 61+ 36V2.. 18. 65— 12-21. 


EQUATIONS CONTAINING RADICALS. 


262. An equation containing a single radical may be 
solved by arranging the terms so as to have the radical 
alone on one side, and then raising both sides to a power 
corresponding to the order of the radical. 


Ex. V27—9+2=9. 
Vz?—9=9—-c. 
By squaring, oe? —9 = 81 — 182 + 2. 
18% = 90. 
Shes oe 


263. If two radicals are involved, two steps may be 
necessary. 


Ex. Vz+15+~V2=15. 


Va +15 + Va = 15. 
Squaring and simplifying, we have 
Va? + 152 = 105 — 2. 
Squaring, we have a? 4+ 15a = 11025 — 210 + a7, 
225 2 = 11025. 
“= 49, 


24 


19. 


20. 


oy SCHOOL ALGEBRA. 


Exercise 90. 


Solve: 

. 2V2+5 = V8. 8. V82+7=8. 

. 8V4e—8=V182—3. 9. 144+V4e2—40=10. 

. Ve+9=5V2 —38. 10. Vl0y—4=Vi7y+11. 
Aone SS. 11. 2V2—2=V82e= oy 
Vb VBy 4. 12. Vip +2=84-Vc. 
.T+2V8a=5. 13. V32+2=16—-V*r. 

. V22—8=—8. 14. Ve—VeH5=V5. 


15. V2z+20—Vx—1—8=0. 
16. Ve+15—T=7—V2—18. 
17) city en 

18. Vx—7T=V2e+1-2. 


Ve—8  Ve+1 we 1+(L—2)t _ 
Ve$3 Ve 8 1-0 


23. Va+Ve+Va— Vi= Vo. 
24. Var —-1=44+1Var—}l. 
25. 8V2 —8Va= Va—Va+ 2Va. 


Oa 
V9+22 








26. V9+22—V2e= 





CHAPTER XVII. 


IMAGINARY EXPRESSIONS. 


264, An imaginary expression is any expression which 
involves the indicated even root of a negative number. 

It will be shown hereafter that any indicated even root 
of a negative number may be made to assume a form which 
involves only an indicated square root of a negative num- 
ber. In considering imaginary expressions, we accordingly 
need consider only expressions which involve the indicated 
square roots of negative numbers. 

Imaginary expressions are also called imaginary numbers 
and complex numbers. In distinction from imaginary num- 
bers, all other numbers are called real numbers. 


265. Imaginary Square Roots. Ifa and 6 are both posi- 

tive, we have 
ee cy Tl (van. 

If one of the two numbers a and 34 is positive and the 
other negative, law I. is asswmed still to apply ; we have, 
accordingly : 

Re 1 Nie V1 = By = 

NAG 5 GaliawWex nol = biv-aL: 

Rats /a ESS AN ab 
and so on. 

It appears, then, that every imaginary square root can 
be made to assume the form aWV—1, where a is a real 
number. 


244 SCHOOL ALGEBRA. 


266. The symbol V—1 is called the imaginary unit, and 
may be defined as an expression the square of which is —1. 


Hence, -V2 1x V— t= (V1 = 
V—ax V—6= Vax V—-1x Vb x V=1 
= Vax Vb x (V—1) 
= Vab x (—1) 
=— Vab. 

267. It will be useful to form the successive powers of 
the imaginary unit. 

(VD) eee es ee eeow eh lee ek kc 

(VE TP EE AO Oe i 

(V—1)' = (V=1)' V=1 = (-1) V-1=- v1; 

(Val) = (VET (VAI =e 

(V—)D§= (V— 1 V-1 = 41) V—-1=4+v-1; 
and soon. We have, therefore, 

(Vr al ee 
(v= Ty" = — 
@/aDtt = =v 
(V—1)"#=+1. 

268. Every nn expression may be made to assume 
the form a+6V—1, where a and 8 are real numbers, and 
may be integers, Sera or surds. 

If 6=O, the expression consists of only the real part a, 
and is therefore real. 


If a=0, the expression consists of only the imaginary 
part 6-V—1, and is called a pure imaginary. 


IMAGINARY EXPRESSIONS. 245 


269. The form a+b~V—1 is the typical form of imaginary 
expressions. 

Reduce to the typical form 6+ V—8. 

This may be written 6+V8xV—l, or G EON Deena: 
here a= 6, and b=2v2. 


270. Two expressions of the form a+ Ves Diab 
are called conjugate imaginaries. 

To find the sum and product of two conjugate imagi- 
naries, 


Ge bv 1 
ey 
The sum is 2a 
Get bee 
GaN aL 
a abv — 1 
—abV—14 2? 
The product is a + 0? 


From the above it appears that the swm and product of 
two conjugate imaginaries are both real. 


271. An imaginary expression cannot be equal to a real 
number. 


For, if possible, let 
atbV—l=e. 
Then transposing a, 6V—1=c—a, 
and squaring, — 6? =(e—ay)’. 


Since 4? and (e—a)* are both positive, we have a nega- 
tive number equal to a positive number, which is impossible. 


946 SCHOOL ALGEBRA. 


272. If two vmaginary expressions are equal, the real parts 
are equal and the imaginary parts are equal. 


For, let a+bV—l=c+dv-1. 
Then (b—d)V—1l=c—a; 
squaring, —(b—d)=(e—ay, 


which is impossible unless 6 =d and a=c. 


278, If x and y are real and x+-yV—1=0, then x=0 
and y=0. 


For, yV—1=—2, 
a y = 2 
a + y = 0, 


which is true only when x=0 and y=0. 


274, Operations with Imaginaries. 


(1) Add 5+ 7V—1 and 8— 9V—1. 


The sum is Ba Br ed a ede 
or RET 


(2) Multiply 8+ 2V—1 by 5—4V—1. 
(S+2V—NG6—4v eu 
= 15 —12V—1+4 10V—1-8(—1) 
O82 i 


(3) Divide 14+ 5V—1 by 2—8vV—1. 
144+5V—1_ (14+5vV—1)(2+3v-1) 
2-3V-1 (2—38V—1)(2+3v—]) 

_13+52V-1 
Ears beet 
_1384+52V—1 

13 
=144V-1. 


IMAGINARY EXPRESSIONS. 247 


Exercise 914. 


Reduce to the form 6V—1: 

ty V9. 9. V— 625. Lye 

2. V—16. 10. V— 36. 18. V= i. 

SPN 25. Wi = G4. Liab 

eee 144 tO 20 20g Oe 

pee 169, 13. */—289. 21. V—(22— By)". 








6. V—2. TAN LO PA N/a 
7. V—8l. 15. > 23. V—(a?+y*). 
8. y/— 256. 16. V—2a°. 24. l= (4), 
Add: 


25. V— 25+ -V— 49 — V— 121. 
age \/ = 64-- V/= 1 — \/— 86. 
a7, V/— at | V—4a' + V— 16a+. 
98.) -/—a? + V—-81a? — V— a’. 
29. a—bV—1+a4+5vV—1. 
30. 24+38V—1—24+8V—1. 
31. d+ b6V—1+e—dv—1. 
32. 8aV—1—(2a—8)V—1. 


Multiply : 
33. V—8 by V—5. 36. V—2? by V—z. 
34. —V/—5 by V—5. 87. V—a by V—7. 


ope — 16 by V— 9. 38. V—8 by V—I6. 


248 SCHOOL ALGEBRA. 

39. V—25 by V— 64. 41. 8V—8 by 2V—2. 
40. V—(a+d) by V—(a—6). 42. —5bV 220 by ox 
43, A/S 2 Biby: Ve eee 
44, pu la as by Pra ssh ss 

2 2 
45. aV=0+8-V=) by 4 ee 
46. 2V—2-48V—3 by 8V—4 = ae 
47. V8+2V—8 by V8 —2V—8. 
48. m—38V—b by n+4V—ce. 


Perform the divisions indicated : 





























Ao ee Sheen 4. 
URGE Vos 38—V—2 

50. ats 5G ee 62. 2+V—=2 
V— 6 V2 1—V-1 

Bite oie 57, Ma. 63, 2a ae 
V4 —V—2 a—aV—1 

py eee ss. Y—8. 64, Oc Meee 
v—8l vV—2 V6—V—8 

53 melts 59. Vv— 102 65. 2a+3bV—1 V =a 
V—b V—52 2a 8b 

54, = ae 60. SV . 66. ta—4bV—1 


Va OG 4a—LbV—1 


CHAPTER XVIII. 
QUADRATIC EQUATIONS. 


275. We have already considered equations of the first 
degree in one or more unknowns. We pass now to the 
treatment of equations containing one or more unknowns 
to a degree not exceeding the second. An equation which 
contains the square of the unknown, but no higher power, 
is called a quadratic equation. 


276. A quadratic equation which involves but one un- 
known number can contain only: 


(1) Terms involving the square of the unknown number. 
(2) Terms involving the first power of the unknown 
number. 
(3) Terms which do not involve the unknown number. 
Collecting similar terms, every quadratic equation can be 
made to assume the form 
ax’ + bza+e=0, 


where a, 6, and c are known numbers, and x the unknown 
number. 

If a, 6, ¢ are numbers expressed by figures, the equation 
is a numerical quadratic. If a, 6, c are numbers represented 
wholly or in part by letters, the equation is a literal quadratic. 


277. In the equation aa*+ bx+e¢=0, a, 6, and ¢ are 
called the coefficients of the equation. The third term ¢ is 
called the constant term, 


250 SCHOOL ALGEBRA. 


If the first power of x is wanting, the equation is a pure 
quadratic; in this case 6 = 0, 

If the first power of x is present, the equation is an 
affected or complete quadratic, 


PURE QUADRATIC EQUATIONS. 


278, Examples, 
(1) Solve the equation 52’— 48 = 22”, 


\ We have 5? —48 = 222, 
Collect the terms, 3 a? = 48, 
Divide by 3, ac} zo 16, 
Extract the square root, v= +4, 


It will be observed that there are two roots, and that these are 
numerically equal, but of opposite signs. There can be only two 
roots, since any number has only two square roots. 

It may seem as though we ought to write the sign + before the 
as well as before the 4. If we do this, we have + x=+4, —z=—4, 
+e=—4 —2=4+4. 

From the first and second equations, 7=4; from the third and 
fourth, e=—4; these values of w are both given by the equation 
w=+4, Hence it is unnecessary to write the + sign on both sides of 
the reduced equation. 


(2) Solve the equation 32?— 15=0. 


We have 3 27 = 15, 
or mies 5, 2 
Extract the square root, r=tvb, 


The roots cannot be found exactly, since the square root of 5 can- 
not be found exactly; it can, however, be determined approximately 
to any required degree of accuracy; for example, it lies between 
2.23606 and 2.23607. 


(3) Solve the equation 32°+ 15=—0. 


We have 32? = — 15, 
or n= — HONE 4 
Extract the square root, a=+V—5, 


QUADRATIC EQUATIONS. ZL 


There is no square root of a negative number, since the square of 
any number, positive or negative, is necessarily positive. 

The square root of —5 differs from the square root of +5 in that 
the latter can be found as accurately as we please, while the former 
cannot be found at all. 


279. A root which can be found exactly is called an exact 
or rational root. Such roots are either whole numbers or 
fractions. 

A root which is indicated but can be found only approx- 
imately is called a surd or irrational root. Such roots 
involve the roots of imperfect powers. 

Rational and surd roots are together called real roots. 

A root which is indicated but cannot be found, either 
exactly or approximately, is called an imaginary root. Such 
roots involve the even roots of negative numbers. 














Spe dare ee le 














Exercise 92. 
Solve: 
827—2=—2°+ 6. 9. Sa F Ete _s, 
5a? +10=62'7+1. 
Ta? — 50 = 4a? 4 25. 10, O@ +8 li-2#_, 
62? -—1L=—42'+4 4. : ° 
: wobihae toa 
5, ett=10. Taek OA BE B 
5 3 4 
327°—8 1 ig cata Spe aly 
6. 10 a4. Fs Be rie eat 9 gy ee . 
7 @a9_ vt ia tpi aed & 
meee ee BR zs—-l «+1 4 
8. 5 ie ie 9 14. PhO pol als At, 


252 
15. 
16. 
17. 
18, 


19. 


20. 


21. 


22. 


23. 
24. 
25. 
29. 


30. 


280. Since (vx + 0? = 2? + 2bx+ 6’, it is evident that 
the expression x? + 26x lacks only the third term, 6’, of 


SCHOOL ALGEBRA. 
82°+1ll2a=1074+8+2?+2. 
(w+ 4)(@+5)=3(@+1)(@+2)—4. 
8(a—2)(e+8)=(e@+1)(e+2) 4245. 
(22 +1)(84—2)+(1—2)(84+ 42) =327 — 15. 
+9 a'—5 , 82?+10 

















Tak eeremsnarars = 
Of DAE Ae pe Oe 

. i 9 ay ==), 

LO ete Ta2k12 eon Die er: 

18 Lior Deg 
gee lee en 26; 04 
@+1l a“—-1 2 i al 
av’?+b=c. 27 tte, oa 








z—a' @2ta 2B 


26 5at26 
28... — + —_ = — 
v+2bx+e=b(2x2+1). aE 32 


25 (x + a)(x+ 6)+ (w@—a)(@ — d)t} =a? +48. 


azvtt-b6 = ba? +a. 


2}(x— a) (a+ 6)+ («@+a)(x—b)} =9a? + 2ab+ B 


AFFECTED QUADRATIC EQUATIONS. 


being a perfect square. 


This third term is the square of half the coefficient of 2. 
Every affected quadratic may be made to assume the form 
a +. 2b2 =c, by dividing the equation through by the co- 


efficient of x”. 


QUADRATIC EQUATIONS. = 0 


To solve such an equation : 


The first step is to add to both members the square of 
half the coefficient of x. This is called completing the square. 

The second step is to extract the square root of each mem- 
ber of the resulting equation. . 

The third step is to reduce the two resulting simple 
equations. 


(1) Solve the equation 2? — 8x = 20. 


We have a? — 8x = 20, 

Complete the square, #?— 8a + 16 = 36. 

Extract the square root, x—-4=+6., 

Reduce, x=4+6=10, 
or a=4—6=—2., 


The roots are 10 and — 2. 


% 


Verify by putting these numbers for # in the given equation 





x= 10, e=— 2, 
10? — 8 (10) = 20, (—2)'—8(—2) = 20, 
100— 80 = 20, 4+ 16 = 20. 
(2) Solve the equation = Dhl Aic eeee 
zx—l e2«+9 

Free from fractions, (w + 1)(# + 9) =(#—1) (4a —3). 
Simplify, OPS 1T eG, 
We can reduce the equation to the form x? — 2b by dividing by 3. 
Divide by 3, ow — lig = 2, 
Half the coefficient of x is 4 of — 1,7 = — 4%, and the square of — 4! 


is 282, Add the square of —1Z to both sides, and we have 


Lix ee 289 
pe hE oud DO) ieee Fae 
cae Tas (a) + 36 
L7\2 361 
2 Jee Se 
or x — 1 =+(7) a 
Extract the root, Sele + 19 


254 SCHOOL ALGEBRA. 


Reduce, pps es 
6 6 
i oti tn done O6 ae 
6 6 6 
17°19 2 1 
or 1 Negi Se —— =e = — 


The roots are 6 and 3 


Verify by putting these numbers for « in the original equation : 








iy == 6. Pee. 

6+1_ 24-3 3 
SEE Te lo 
7 _ 21 ae 
those l cat _i 9 

3 37 

2 

4 26 

Exercise 93. 
Solve: 


Lee Dee. 3. 5. 2? + 5e= 14. 9. # +22 =40, 
2. 7—624=7. 6. 2 —38x= 28. 10. 82°—42=4. 
Br A La ee ees ll. 62° -- eae 
4, 2°+4x2=5. 8. 52° + 8% =2: 12. 627:->@ ae 

18. 1227;—1llz+2=0. 

14. 1527—2x—1=0. 

15. (7 +1 G42) _ Ga DG~)—s 


16, 2% 3)e_ (@+4)@—1)_ 47 
4 6 


QUADRATIC EQUATIONS. 255 























ox+5 , 2a—5 et+2 4-—a“_7 
17. = 8, 7 fi tn la a eaten 
6-4. : E> 2 el Qa 63 
Pee FO _ 1. 93, Tr8_oe+8 
a—2 22+1 x—-2 -2+4 
19. pret 1+ ae _ 9. 24. LEE tile sao 
Ait l—x 2 3 2x2—1 
x z+1. 18 gt+1,2#2+2 18 
20. ——- + —— = —. 25. — oo 
peela x 6 Soe a 6 
Sloe —4¢=— 1. 26. T2?—82=—1. 


ANOTHER METHOD oF CoMPLETING THE SQUARE. 


281. When the coefficient of «? is not unity, we may 
proceed as in the preceding section, or we may complete 
the square by another method. 

Since (aa + 6)? = aa? + 2abx + 6’, it is evident that the 
expression a7? + 2abzx lacks only the third term, b’, of being 
a complete square. 

It will be seen that this third term is the square of the 
quotient obtained from dividing the second term by twice the 
square root of the first term. 


282. Every affected quadratic may be made to assume 
the form of aa’? + 2abx =e. 


To solve such an equation: 


The first step is to complete the square; that is, to add to 
each side the square of the quotient obtained from dividing 
the second term by twice the square root of the first term. 

The second step is to extract the square root of each side 
of the resulting equation. 

The third and last step is to reduce the two resulting 
simple equations. 


9256 SCHOOL ALGEBRA. 


(1) Solve the equation 162’ + 52—3=72?—2x-+ 46. 


We have 1627 +5¢%—3=72?—2 +46." 
Simplify, 92? +62= 48. 


To complete the square, take the square root of 947. This is 32, 
and twice 3x is 6a. Divide the second term by 6a, and the quotient 
is 1. Square 1 and add it to both sides. 

9a? +64%+4+1=49. 
Extract the square root, 3e+1=+7. 
Reduce, 3¢=—1+7o0r—1—-7, 
32 =6 or —8. 
*, @©= 2 or — 22, 
Verify by substituting 2 for x in the equation: 
162? + 54¢—3=72? —2 + 45, 
16 (2)? + 5 (2) — 3 = 7 (2)? — (2) + 45, 
64+10—3 = 28 —2+ 45, 
ih il 
Verify by substituting — 22 for # in the equation: 
1647 ++5e¢—3=72? —2 + 45, 


16(—3) + o( =F) se 7(=5)= (=3) 445; 


1024 40 448 8 
on eee Lee oe 
Se aed ea 

1024 — 120 — 27 = 448 + 24 + 405, 

877 — B77. 


(2) Solve the equation 32?—42= 82. 


Since the exact root of 3, the coefficient of 2?, cannot be found, it 
is necessary to multiply or divide each term of the equation by 3 to 
make the coefficient of «? a square number. 


Multiply by 3, 9a? — 12% = 96. 
Complete the square, 
9a?-—124+4+4=100. 
Extract the square root, 3a—2=+ 10. 
Reduce, 32=2+10 or 2—10; 
3% = 12 or —8. 
*, ©=4 or — 22, 


- 


QUADRATIC EQUATIONS. 257 
Or, divide by 3, 


Complete the square, 2? = 





O53 oer oe: 
Extract the square root, «— : =+ : 
2+10 
c= ’ 
3 
= 4 or — 22. 


Verify by substituting 4 for x in the original equation : 


48 — 16 = 32, 
32 = 32. 
Verify by substituting — 22 for x in the original equation: 
214 — (— 102) = 32, 
32 = 32. 
(3) Solve the equation — 327+ 54 = — 2. 


Since the even root of a negative number is impossible, it is neces- 
sary to change the sign of each term. The resulting equation is 
3.0% — 5x = 2. 


Multiply by 3, 9a? — 15% =6. 


Complete the square, 





Sifted 
5 7 
Extract the square root, 3a— ae 
Reduce, 352 ; 7 
32=6 or —1 
x = 2 or —= 
Or, divide by 3, a2 — “ = 2 
Complete the square, 
2 — 5a ve 25 ai 49 


258 SCHOOL ALGEBRA. 


Extract the square root, «— : =+ z 
pee we 
6 
= 2or— L 
3 


284, If the equation 32?— 5x = 2 be multiplied by four 
tumes the coefficient of x”, fractions will be avoided. 
We have 36.2? — 60a = 24, 
Complete the square, 
36 a? — 604 + 25 = 49. 
Extract the square root, 6a —5=+ 7. 


62=5+4 7. 

62=12 or —2. 

# ete en 
3 


It will be observed that the number added to complete the square 
by this last method is the square of the coefficient of x in the original 
equation 327 — 52 = 2. 

Nore. If the coefficient of x is an even number, we may multiply 
by the coefficient of «?, and add to each member the square of half the 
coefficient of x in the given equation. 


(1) Solve the equation 42?— 23% = — 30. 


Multiply by four times the coefficient of 2, and add to each side 
the square of the coefficient of a, 


64 2? —() + (23)? = 529 — 480 = 49. 
Extract the square root, 8#—23 = + 7. 


Reduce, 8a=2347; 
8a = 30 or 16. 
*, e = 32 or 2. 


Nors. Ifa trinomial is a perfect square, its root is found by taking 
the roots of the first and third terms and connecting them by the sign 
of the middle term. It is not necessary, therefore, in completing the 
square, to write the middle term, but its place may be indicated as 
in this example. 


QUADRATIC EQUATIONS. 259 


(2) Solve the equation 722? — 30% = — 7, 


Since 72 = 2° x 3?, if the equation is multiplied by 2, the coefii- 
cient of x? in the resulting equation, 1442?— 60a =— 14, will be a 
square number, and the term required to complete the square will be 


( syle Ge =>. Hence, if the original equation is multiplied by 


4 x 2, the coefficient of x? in the result will be a square number, and 
fractions will be avoided in the work. We shall then have 


576 2? — 2402 = — 56. 
“, 5762? —() +25 =—81. 
Extract the root, WMge5=tvV—3l, 
w= 3 (5+ V— 831). 


Exercise 94. 


Solve: 
omer — 6. 14. Ba? +22 = 26, 
ooo. — os = 27. 3 
2 =a 
ese a, — 5 157 a =p ae ee 
4. 2a? —5a=7. 16, 2 —2—2(e—2) 
5, o2'e-iz—6 ip is > 
x 
6. 5a?—Tx = 9A ht ete tora 
if 827+ 32 = 26. 18. OO to a3 say k's 
16 
8. 12° +52 =150. 20 
19. 32°+ 32 =—- 
9. 67+57=14. att 
10. 72°24 =$. Coie a 
Preor + iz= 51. 21. wo? 
xv 
aie ons 
13677 — 202 = 75. ob . (8+ 210 


13. llz?—10x2 = 24. mini ions ae 


260 SCHOOL ALGEBRA. 
23. («+ 2)(2¢+ 1): (Cos) (Ben eee 
24. 3x2(22+5)—(#+ 38)(84%—1)=1. 
ob. Cet eT) eCe FD 5, 


26. 2(52° —8x—6)—4 (2? — 3) =22+1. 




















ites eee, $0, 
+3 2 x—-l «#-—-2 x«—-—4 
Beha e ste eee 31, tt 2 ee 
x—l 2x4+1 838 x—-4 2-2 
7 + 24—3 , 5—38-2 
29. ———. + ———_= 9. oak mes be 
Te OW Om ee ; 4 
33. 11— 8a, 2744) _ 1 


l—=z 1—22 


etl, l—#@ 2 


34. == . 
— 4-208 5 (a — 2) 















































35. 
eer A z+] 

of AE 3 Sled ee ek eee 

" 9Qe—1) Q2e+l) 4—82 
22—1 3 x—2 

377. ep eet 5 

A 3 x—8 peer 

38. 382+2 Ca el 
2a—l 2¢4+1 427-1 
—5 ,x2-—8 80 1 

39) = = 
xt+3 «2-8 e912 

40. Zepl, ser) _ 45 eds 


1—2 T+tae 49-2 


QUADRATIC EQUATIONS. 261 


LITERAL QUADRATICS. 
285. Examples. 
(1) Solve the equation az’+ br+e=0. 


Transpose ¢, ax* + be=— ce. 
Multiply the equation by 4a and add the square of 8, 
47a? +()+ 0 =0? —4ac. 


Extract the root, 2ax+b=+ Vb? —4ac. 
° eae VE Aue 
a 2a 


(2) Solve the equation (a? + 1)2 = az? +a. 
Transpose aa? and change the sign, 
ax* — (a? + 1)ex=—a. 
Multiply by 4a, and complete the square, 
4 atx? —()+(a?+1P?=—4a? +at+ 20? +41 


= at — 2a? +1. 
Extract the root, 2a%—(a?+1)= + (a?—1). 
Reduce, 2aux = (a? + 1) + (a? — 1), 
= 2a? or 2. 
: 1 
. £=¢@ or —- 
a 


(3) Solve the equation adx — aca? = bex — bd. 


Transpose bca and change the signs, 

acax* + bex — adx = bd. 
Express the left member in two terms, 

acu? + (be — ad) a = bd. 
Multiply by 4ac, and complete the square, 
Aatcta? + () + (be— ad)? = bc? + 2abced + a?d?, 

Extract the root, 

2acx + (be — ad) = + (be + ad). 


Reduce, 2acuw = — (be — ad) + (be + ad) 
= 2ad or — 2be. 
here! Ret 
. == or—-=: 


c a 


262 SCHOOL ALGEBRA. 





(4) Solve the equation pa? — pet qu + gx pe 
Express the left member in two terms, 

(p + 9)2?—(p—g)e=— 
Multiply by four times the coefficient of 2, 

4(p + qa? — 4(p* — q°) «= 4g. 
Complete the square, 
4(p + gPat—() + (p— 9) =p + 2pg + g. 

Extract the root, 

2(p -+.9)2—(p—g)—=(p a gy 


Reduce, 2(p+qe=(p—gQ+(pt+q), 
= 2p or — 2g. 

giant. Se 

P+ Pt+q 


Notre. The left-hand member of the equation when simplified 
must be expressed in two terms, simple or compound, one term con-. 
taining «?, and the other term containing a. 


Exercise 95. 





Solve: 
1. a7 -+ 2ar=3a'. 9. 202? + ax —1=0. 
Ott A = at 10. 120°x?— 5brx4=3. 
3. a?+8be = 98%. 11. 22 4 211 a(v—8a). 

re ete 2 
4. 2°+3b2 =100'. #5 Bat eee 
5. 27+ 5axr = 140’. pes ie ey ait, 
6. 827+ 4er = 4’. 13. yi & 8S 

a 4@ 

: — 227? = 2a’. 
Teor x a Sae* , 20 _ 18 
8. 62°?—ar—av=0. i et 3 a 


2 2 ON Te 
15. epee + SHE Hae, 


QUADRATIC EQUATIONS. 263 








19. 


20. —— 


21. ——— 


37. 
38. 
39. 
40. 


29. 
30. 
31. 
32. 
33. 
34. 


35. 


36. 








ot, e+ a= 2a’, 22. 2+ (a—b)x=ab. 

peg) e _-. os 2G, Oa 8 
ataz 3 2a—x at+2uz 8 
a(de—a)_ a oA 22—3a , 38a+2a_10 

+ ae 12 apc 4z—a 7 
ee «gg Bong 
wer mn icra ' 68? Gab at 
xta Bate 5 26. Ga Be ay bus 
b—a ta atbtez «+6 
a+2b,a+6b_5 1 [ieee Ce | 

= PW ie =-+—+-. 
| ay atb4-z2 Pee va 
28. Pet a Oat. 


9a? — 8(a42b)¢4+ 2ab=0. 
(2a+1)27?+3a2+a—a’=0. 
fea 2(1+a’)x+1l—a’=0. 
(a+ 6)? — (a? — 0°) x= ab. 

ae? + 2bert a= cx? +2a%r+ 0. 
(a+ 6)x?—(2a+6)r4+a=0. 


2e—36 _ 
a—26 


DO Ly RW 
z2+2a 2(a—b) 





x—b 


(8a? + 0?) (2? —x+1)=(?+380")(¢’+2+1). 
2 —(a+ b)x+ac+ be—e?=0. 
a —p2=(p+q+r)(q+7). 


(a? + ab) 


(2? —1)—-(¢+0’)2+(a+b)(a—26)=0. 


264 SCHOOL ALGEBRA. 


SOLUTIONS BY FACTORING. 


286. A quadratic which has been reduced to its simplest 
form, and has all its terms written on one side, may often 
have that side resolved by znspection into factors. 

In this case, the roots are seen at once without com- 
pleting the square. 


(1) Solve 27+ 7%—60=0. 


Since a* + 72 — 60 = (a + 12) (a —5), 
the equation x? + 7a—60=0 
may be written (x + 12)(a—5)=0. 


If either of the factors x +12 or «—85 is 0, the product of the two 
factors is 0, and the equation is satisfied. 
Hence, x+12=0, and r—5=0, 
* e=—12, and «=5., 


(2) Solve 2a°—2?—6x=0. 

The equation 223 — a? -—62=0 
becomes x (2a? —a«2—6)=0, 
and is satisfied if «#=0, or if 2227—-x%—6=0, 


By solving 2x? — x —6=0, the two roots 2 and — 3 are found. 


Hence the equation has three roots, 0, 2, — . 


(3) Solve a+ 2?—42 —4=0. 
The equation oe +a2—42-—4=0 
becomes a (a +1)—4(e@ + 1)=0, 
(a? ~ 4)(a + 1) =0. 
Hence the roots of the equation are —1, 2, — 2. 


(4) Solve 2° — 22?—11lz+12=0. . 
By trial we find that 1 satisfies the equation, and is therefore a 
root (¢ 89). 
Divide by «—1; the given equation may be written 
(a — 1) (x? — x —12)=0, 
and is satisfied if e —1=0, or if 2?—2—-12=0, 
The roots are found to be 1, 4, —3. 


QUADRATIC EQUATIONS. 265 


(6) Solve the equation x(a — 9) =a(a@ — 9). 
If we put a for x, the equation is satisfied; therefore a is a 
root (¢ 89). 
Transpose all the terms to the left-hand member and divide by 
a. ; 
The given equation may be written 
(aw — a) (a? + ax + a? — 9) =0, 
and is satisfied if s—a =0, 
or if 227+ ax+a?—9=0. 
The roots are found to be 
ices V36—3a? —a— V36—3a? 
2 2 


Exercise 96. 


Find all the roots of 


1. 2#—5xe+4=0. 5. a +2°7—62=0. 
2. 627—52—6=0. 6. 2 —8=0. 
8. 227?—zx—38=0. eae 8 = 0: 
4. 102°?+2—3=0. 8. t 160; 


9. («—1)(x—38)(#’+52x+6)=0. 

10. (2% —1)(«#— 2) (82°—52—2)=0. 

11. (a? +2 — 2)(22?+32—5) =0. 

12. 2 +a?—4(7+1)=0. 

13. 32°+4 22?— (82+ 2) =0. 

14. 2° —27—182+39=0. 
15. 2484 3(2?—4)=0. 17. 22°—22°—(2’?—1)=0. 
16. x(2?—1)—6(a—1)=0. 18. ae —3x—2=0. 

19. 2a°+ 227+ (2? —5x2—6)=0. 

20. 2*—42°—2?+162—-12=0. 


266 SCHOOL ALGEBRA. 


SoLUTIONS BY A FORMULA. 


287. Every quadratic equation can be made to assume 
the form az’?+ ba+ce=0. 
Solving this equation (§ 285, Ex. 1), we obtain for its 
two roots 
—b1ivVb—4ac —b= Vea 4ac 
2a 2a 


There are two roots, and but two roots, since there are 
two, and but two, square roots of the expression 6’— 4ac. 

By this formula, the values of # in an equation of the 
form az’-+ bz -+-¢=0 may be written at once. 


Ex. Find the roots of the equation 8z?—5%+2=0. 
Here a=8, b=—5, c=2. 
Putting these values for the letters in the above formulas, we have 


_5+V25—24 9. 5— V25—24 


6 6 
= or 4 
=1 or 2. 
Fixercise 97. 
Solve by the above formulas: 
1,22) oa= 14, 7 52—Ta=—2. 
2.00 — ox = 12, 8. 42° — 9x = 28. 
Stag ie 38: 9. 5a? -+ fa male 
4. 5a?—2 = 42, 10. Mat —92 = 
5. 62°—T7x2=10. 11. 72?+52=88. 
6. 3a7—lla=—6. 12.5379 Tan 


QUADRATIC EQUATIONS. 267 


EQUATIONS IN THE QUADRATIC Form. 


288, An equation is in the quadratic form if it contains 
but two powers of the unknown number, and the exponent 
of one power is exactly twice that of the other power. 


289, Equations not of the second degree, but of the 
quadratic form, may be solved by completing the square. 


(1) Solve: 82° + 632° = 8. 


This equation is in the quadratic form if we regard a? as the un- 
known number. 
We have 8 x8 + 63 a3 = 8. 
Multiply by 32 and complete the square, 
256 a® + ( ) + (63)? = 4225. 
Extract the square root, 1623 +63 = +65, 
Hence, a8 = : or — 8. 


Extracting the cube root, two values of w are } and —2. There 
are four other values of « which we do not find at present. 


(2) Solve: Va 28 4/ a = 40. 
Using fractional exponents, we have 
a? — 3a = 40, 


This equation is in the quadratic form if we regard a as the un- 
known number. 


Complete the square, 4a?—12a3 +9 = 169. 


Extract the root, G4 3 = 13, 
2x% = 16 or — 10, 
re a8 of 6, 0 


x = 16 or —5 V5. 


There are other values of « which we do not find at present. 


268 SCHOOL ALGEBRA. 


(3) Solve completely the equation z*°= 1. 


We have a —1=0. 
Factoring, («—1)(#+a+1)=0. 
Therefore, either z—-1=0 
or alee or a oa Ge 
z—1=0. aw? +a2+1=0, 
owen ld, eae ey 
: Solving, #= aw. 
The three values, 1, alts ge rt haat are the three cube 
roots of 1. et 
(4) Solve: (24% — 3) — (2% — 38) = 6. 
Put y for 2% — 3, and therefore y? for (2% — 3). 
We have y?—y =6. 
Solving, y =3 or —2. 
Putting now 2a—8 for y, 
2%—3 = 3, 24—3=— 2, 
x= 3. v=}. 


Exercise 98. 


1. af —5277+4=0. 6. 10z*—2L=a7 

2. 2'— 1827+ 36=0. 7. Vei+8Ve= 18 

3. 2 — 212? =100. 8. 8Vx2—2Vx = — 20. 

4. 42°— 327° = 27. 9. 54"+32"= 68. 

5. Qat+ 5a? = 218, 10. (82+3)*+ (8x48) =380. 
11. 2(2?—2+1)—V2?—-2+1=1. 

12. 2°—92°+8=0. 14. (e+1)4+ Vat TG: 
af el 

13. «+2 =e 15. 2*— 182? =— 36, 


16. 2274+4274+94 8V22?4+ 44+9=40. 
122° — 1127+ 102 — 78 1 


17. mee ee 
82° —Ta+6 ag 


QUADRATIC EQUATIONS. 26 


CO 


RADICAL EQUATIONS. 


290. If an equation involves radical expressions, we first 
clear of radicals as follows: 


Solve V2 +44 V22+6=Vizc+l4. 

Square both sides, 
e+4+2V(e+4j(22+6)+204+6=7x7 +14. 

Transpose and combine, 2V(«#+4)(2a¢+6)=4a+4+4. 


Divide by 2 and square, (7 + 4)(2” + 6) = (24+ 2)% 
Multiply out and reduce, x? — 3a = 10. 
Hence, x=5or—2. 


Of these two values, only 5 will satisfy the original equation. 
The value — 2 will satisfy the equation 


Vi+4—V20+6=Vi0+ 14. 
In fact, squaring both members of the original equation is equiva- 
lent to transposing \/7z + 14 to the left member, and then multiplying 


by the rationalizing factor Vz +44+ V22+6+ V7 +14, so that 
the equation stands 


(Ve +44+ V204+6—V704+14)(V24+44 V204+6+4+V70+4+14)=0, 
which reduces to V(a + 4)(2a + 6)—(2a + 2)=0. 
Transposing and squaring again is equivalent to multiplying by 
(Va +4~—V20+6 4+ V724+14)(ve+4-— V2046~Vi724 14). 
Multiplying out and reducing, we have 
x? —3x%—10=0. 
Therefore, the equation a? —3«2—10=0 is really obtained from 
(Va +44+ V204+6-—Vi72+14) 
x (V2 +44 V204+6+4+ V7a2 414) 
x (Wa +4-— V2e4+6— Via +14) 
x (Vot4d— Vie tb + Vin +14) =0. 
This equation is satisfied by any value that will satisfy any one 
of the four factors of its left member. The first factor is satisfied 


by 5, and the last factor by — 2, while no values can be found to 
satisfy the second or third factor, 


270 SCHOOL ALGEBRA. 


Hence, if a radical equation of this form is proposed for solution, 
if there is a value of # that will satisfy the particular equation given, 
that value must be retained, and any value that does not satisfy the 
equation given must be rejected. (See Wentworth, McLellan and 
Glashan’s Algebraic Analysis, pp. 278-281.) 


291, Some radical equations may be solved as follows: 


Solve 72? —5274+8V727—57+1=—8. 
Add 1 to both sides, 

"et? —5a+148V7e—5e24+1=—7. 
Put V7a?—5a+1=y; the equation becomes 





y+ 8y=—7. 
Hence, y=—lor—7, 
y? = 1 or 49. 


We now have 72?—52+4+1=1, or 72?—524+1=49. 
Solving these, we find for the values of a, 
Baht at 
7 7 
These values all satisfy the given equation when we take tle 
negative value of the square root of the expression 727—5a+1; 


they are in fact the four roots of the biquadratic obtained by clear- 
ing the given equation of radicals. 


0 


Exercise 99. 
Solve: 


1. V92+404+ 2Vet+7= V2. 
2. V Gee Va—x2= vb. 


82+ V4r—2" _» 
8a—WV42—2? 


4. Vz—38—WV2—14= V4a— 155. 
os Vae+4—Va=Va+ 8. 


QUADRATIC EQUATIONS. 


8Ve—4 154+38V2_ 
QtVe 404+-~Vz 


V142+9+2V2+14V32+1=0. 


Voth oN et 1 =), 
Ve—234+V2+3—V4ze2+1=0. 


ey ie 1 A/8e 10 ino 8 = 0: 
oy Yd + Vel + 8/5141 =0. 
See SV Oe | a + 2. 

. V2+2—V2—2—V2e=0. 


1 i 


——_______.. + ——______ = 12, 
SY 2 ee i eee 


Mysore piso. 


V/32+13 


See Ve oe —2=0. 


824—~vV2?—8 


es 
Co HS Ea 





ae boa 1 Sa 9S 8: 
soe + ihe — 229 52 1 = 2 

ee 89a 82 oO 7, 

Bigg e/g ee 8 A gO. 

. 8a— 424+ V382°—42—6=18. 
Peat ONO e — 160 21162. 


972 SCHOOL ALGEBRA. 


292. Problems involving Quadratics. Problems which in- 
volve quadratic equations apparently have two solutions, 
since a quadratic equation has two roots. When both roots 
of the quadratic equation are positive integers, they will 
give two actual solutions of the problem. 

Fractional and negative roots will in some problems give 
admissible solutions; in other problems they will not give 
admissible solutions. ? 

No difficulty will be found in selecting the result which 
belongs to the particular problem we are solving. Some- 
times, by a change in the statement of the problem, we may 
form a new problem which corresponds to the result that 
was inapplicable to the original problem. 

Imaginary roots indicate that the problem is impossible. 

Here as in simple equations x stands for an unknown 
number. 


(1) The sum of the squares of two consecutive numbers 
is 481. Find the numbers. 


Let x = one number, 
and x +1 =the other. 
Then x? + (# + 1)? = 481, 
or 207+ 2%+1=481. 


The solution of which gives «= 15 or —16. 

The positive 15 gives for the numbers, 15 and 16. 

The negative root —16 is inapplicable to the problem, as consecu- 
tive numbers are understood to be integers which follow one another 
in the common scale, 1, 2, 3, 4..... 


(2) A pedler bought a number of knives for $2.40. 
Had he bought 4 more for the same money, he would have 
paid 8 cents less for each. How many knives did he buy, 
and what did he pay for each? 


Let x = number of knives he bought. 


Then ae = number of cents he paid for each. 


QUADRATIC EQUATIONS. 273 





But if xz + 4= number of knives he bought, 
240 _ number of cents he paid for each, 
x +4 
240 __240 _ the difference in price. 
we 2+4 
But 3 = the difference in price. 
. 240 240 _. 
“Np ee ee ea 
Solving, z= 16 or — 20. 


He bought 16 knives, therefore, and paid 242, or 15 
cents for each. 
If the problem is changed so as to read: A pedler bought 
a number of knives for $2.40, and if he had bought 4 dess 
for the same money, he would have paid 3 cents more for 
each, the equation will be : 
Fath Bet 
e—-4 & 
Solving, x = 20 or — 16. 
This second problem is therefore the one which the neg- 
ative answer of the first problem suggests. 


(3) What is the price of eggs per dozen when 2 more in 
a shilling’s worth lowers the price 1 penny per dozen? 





Let « = number of eggs for a shilling. 
Then >= cost of 1 egg in shillings, 
and 12 cost of 1 dozen in shillings, 
But if ay i = number of eggs for a shilling, 
12 — cost of 1 dozen in shillings. 
x+2 
ee ea (1 penny being 7; of a shilling). 


eo @¢+2 12 


The solution of which gives = 16, or — 18. 
And, if 16 eggs cost a shilling, 1 dozen will cost 9 pence. 


Therefore, the price of the eggs is 9 pence per dozen. 


274 SCHOOL ALGEBRA. 


If the problem is changed so as to read: What is the 
price of eggs per dozen when two /ess in a shilling’s worth 
raises the price 1 penny per dozen? the equation will be 


ae de eek 





The solution of which gives «= 18, or —16. 
Hence, the number 18, which had a negative sign and was inappli- 
cable in the original problem,-is here the true result. 


Exercise 100. 


1. The sum of the squares of two consecutive integers is 
761. Find the numbers. 


2. The sum of the squares of two consecutive numbers ex- 
ceeds the product of the numbers by 18. Find the numbers. 


3. The square of the sum of two consecutive even num- 
bers exceeds the sum of their squares by 336. Find the 
numbers. 


4. Twice the product of two consecutive numbers ex- 
ceeds the sum of the numbers by 49. Find the numbers. 


5. The sum of the squares of three consecutive numbers 
‘is 110. Find the numbers. 


6. The difference of the cubes of two successive odd 
numbers is 602. Find the numbers. 


7. The length of a rectangular field exceeds its breadth 
by 2 rods. If the length and breadth of the field were 
each increased by 4 rods, the area would be 80 square rods. 
Find the dimensions of the field. 


8. The area of a square may be doubled by increasing 
its length by 10 feet and its breadth by 3 feet. Determine 
its side. 


QUADRATIC EQUATIONS. aT 


9. A grass plot 12 yards long and 9 yards wide has a 
path around it. The area of the path is 2 of the area of the 
plot. Find the width of the path. 


10. The perimeter of a rectangular field is 60 rods. Its 
area is 200 square rods. Find its dimensions, 


11. The length of a rectangular plot is 10 rods more 
than twice its width, and the length of a diagonal of the 
plot is 25 rods. What are the dimensions of the field? 


12. The denominator of a certain fraction exceeds the 
numerator by 8. If both numerator and denominator be 
increased by 4, the fraction will be increased by 2. Deter- 
mine the fraction. 


13. The numerator of a fraction exceeds twice the de- 
nominator by 1. If the numerator be decreased by 3, and 
the denominator increased by 3, the resulting fraction will 
be the reciprocal of the given fraction. Find the fraction. 


14. A farmer sold a number of sheep for $120. If he 
had sold 5 less for the same money, he would have received 
$2 more per sheep. How much did he receive per sheep? 

State the problem to which the negative solution apples. 


15. A merchant sold a certain number of yards of silk 
for $40.50. If he had sold 9 yards more for the same 
money, he would have received 75 cents less per yard. 
How many yards did he sell? 


16. A man bought a number of geese for $27. He sold 
all but 2 for $25, thus gaining 25 cents on each goose sold. 
How many geese did he buy? 


17. A man agrees to do a piece of work for $48. It 
takes him 4 days longer than he expected, and he finds 
that he has earned $1 less per day than he expected. In 
how many days did he expect to do the work? 


276 SCHOOL ALGEBRA. 


18. Find the price of eggs per dozen when 10 more in 
one dollar’s worth lowers the price 4 cents a dozen. 


19. A man sold a horse for $171, and gained as many 
per cent on the sale as the horse cost dollars. How much 
did the horse cost? 


20. A drover bought a certain number of sheep for $160. 
He kept 4, and sold the remainder for $10.60 per head, and 
made on his investment 3 as many per cent as he paid dollars 


for each sheep bought. How many sheep did he buy? 


21. Two pipes running together can fill a cistern in 52 
hours. The larger pipe will fill the cistern in 4 hours less 
time than the smaller. How long will it take each pipe 
running alone to fill the cistern? 


22. A and Bean do a piece of work together in 18 days, 
and it takes B 15 days longer to do it alone than it does A. 
In how many days can each do it alone? 


23. A boat’s crew row 4 miles down a river and back 
again in 1 hour and 30 minutes. Their rate in still water 
is 2 miles an hour faster than twice the rate of the current. 
Find the rate of the crew and the rate of the current. 


24. A number is formed by two digits. The units’ digit 
is 2 more than the square of half the tens’ digit, and if 18 
be added to the number, the order of the digits will be 
reversed. Find the number. 


25. A circular grass plot is surrounded by a path of a 
uniform width of 8 feet. The area of the path is % the area 
of the plot. Find the radius of the plot. 


26. Ifa carriage wheel 11 feet round took 4 of a second 
less to revolve, the rate of the carriage would be 5 miles 
more per hour. At what rate is the carriage travelling? 


CHAPTER XIX. 


SIMULTANEOUS QUADRATIC EQUATIONS. 


298, Quadratic equations involving two unknown num- 
bers require different methods for their solution, according 
to the form of the equations. 


Case I. 


294, When from one of the equations the value of one of the 
unknown numbers can be found in terms of the other, and this 
value substituted in the other equation. 


Ex. Solve: 82° —Qaey=5 \ (1) 
z—y=2 (2) 
Transpose x in (2), y =x — 2. 


In (1) put « —2 for y, 
3a? — 2a(x% — 2) =5, 


The solution of which gives  =1, or «= —5. 
If 2=1, 
y=1—-2=-1 
and if x= — 5, 


y=—5-—2=—7, 
We have therefore the pairs of values, 
fee : Ele ¥ 
The original equations are both satisfied by either pair of values. 


But the values «= 1, y=—7, will not satisfy the equations; nor will 
the values «=—5, y=—1. 


The student must be careful to join to each value of x 
the corresponding value of y. 


278 SCHOOL ALGEBRA. 


Case II. 


295. When the left side of each of the two equations is 
homogeneous and of the second degree. 





Solve: Rae (1) 
y?— 2? = 16 (2) 
Let y = va, and substitute vx for y in both equations. 
From (1), 2072? — 4 ve? + 327 =17, 
“3 17 
2. 02 = ———____.. 
2v?— 4043 
From (2), va? — xo = 16. 
ge 
v—1 
17 16 





Equate the values of a?, yh eae ET. 


32v? — 640 +48 =17r? —17, 
15? — 64v = — 65, 
225 v? — 960 = — 975, 
225 v? — ( ) + (32)? = 49, 


159 = 32 = +7; 
5 13 
.v== or —: 
5 5 
If eo) If v8 
se 5 
y 3 5 
Substitute in (2), Substitute in (2), 
95 op 2 16 169 x “ds a2 = 16, 
Ce h perme st 25 
ff == 9. a? — 2, 
e=+3, 9 
5 r= Fei 
y= 1 = +5, 3 
sag gis 
d= 5 3 


SIMULTANEOUS QUADRATIC EQUATIONS. 279 


Case III. 


296. When the two equations are symmetrical with respect to 
x and y; that is, when x and y are similarly involved, 


Thus, the expressions 
20°4327y?+2y', 2xy—8x—3y4+1, e—8axy—38ay+y', 


are symmetrical expressions. In this case the general rule 
is to combine the equations in such a manner as to remove 
the highest powers of x and y. 


Solve: at + x! = 387 ; (1) 
ZY 4 (2) 


To remove 2‘ and 74, raise (2) to the fourth power, 


at + 4a3y + 6a7%y? + 4ay?+ yt = 2401 
Add (1), af + yt= 337 


Dat + dady + Gary? + Lay? + 2yt = 2738 
Divide by 2, at + 2a%y + 3a%y? + 2ayi+ y* = 1369. 
Extract the square root, e+ ay + y? = + 37. (3) 
Subtract (3) from (2)?, wy = 12 or 86. 
We now have to solve the two pairs of equations, 


a py= 7). et+y= 7). 








Ly = 12 ; xy = 86 
From the first, fps inet = 3 
y=3 y=4 
From the second, PERS. i. 295 
7*V — 295 
y= 


280 SCHOOL ALGEBRA. 


297. The preceding cases are general methods for the 
solution of equations which belong to the kinds referred to; 
often, however, in the solution of these and other kinds of 
simultaneous equations involving quadratics, a little inge- 
nuity will suggest some step by which the roots may be 
found more easily than by the general method. 


(1) Solve: a+y=40 \ (1). 
xy = 800 (2) 
Square (1), x + 2ay + y* = 1600. (3) 
Multiply (2) by 4, 4 ay = 1200. (4) 
Subtract (4) from (3), 
x — 2ey + y* = 400. (5) 
Extract root of each side, «—y = + 20. (6) 
From (1) and (6), 30S eet Ae 


Psi oey 
2) Solve: = | == (1) 
(2) Solve aa Fe DG 

Lee oes 1 

ay 400 ©) 
3 biel 
Square (1), caress: evs (3) 


Subtract (2) from (3), 


| 


Slr Sle 
aa 


4 
Subtract (4) from (2), 


pet nee ee 
mt ey ~ ay? -400 
Extract the root, aed (5) 
oY 20 
From (1) and (5), ae or te 
y=5)'  y=4 


SIMULTANEOUS QUADRATIC EQUATIONS. 


(3) Solve: 


Square (1), 


Subtract (2) from (3), 


x—y= 4 

x+y? = 40 

x? —2ey+y?=16. 
— 2ay = — 24. 


Subtract (4) from (2), 


Extract the root, 
From (1) and (5), 


(4) Solve: 


Divide (1) by @), 


Square (2), 


Subtract (3) from (4), 


Divide by —3, 
Add (5) and (3), 
Extract the root, 
From (2) and (6), 


(5) Solve: 


Divide (1) by (2), 


Square (2), 


Subtract (4) from (3), 


which gives 


We now have, 


Solving, we find, 


a + 2ey + y? = 64. 
ety=+8. 
x“ =='6 


. z=—2 . 
yas 


y=—6 


me chet 
Dey ==) oT 
a — xy + y? = 13. 
a + 2ay + y? = 49. 
dvy = 36. 
—ay=—12. 
a? —2ey+y?=1. 


e+y =12 
ot ay + y? = a. 


w+ 2ey + y? = 144. 
xy = 32. 


ety=12 : 
xy = 32 


aga 


281 


(1) 
(2) 


(3) 
(4) 


(5) 


(1) 
(2) 


(3) 
(4) 


(5) 


(8) 


(1) 
(2) 


(3) 
(4) 


10. 


gia Be 


12. 


13. 


14. 


15. 


16. 


17. 


SCHOOL ALGEBRA. 


Exercise 101. 


ety= } 3 patie 5. ee: 
zy == 10 ry =—8 v+y = 80 
aaa 4 ad iat 6 «ty = a 
xy = 27 vy = 11 e+ y? = 29 
2—-y= = 18. Be 
xv’ +-y? = 45 3842—y=s3 
eos ae 19) 
V+ Y= os 7) ee 

Gf = 16 aN Baa 
5a —4y = 10 at ae 
82°—4y= 8 oy ee 
ae ant 21. ete=2 | 
ry = 6 pet ad 
22 —By =2 y=6 | 

2 2ay=—T 

es a2, -+>—=11 
2% —3y = at oy nae 

3x" — 4ay = 32 1,1 _6) 

x’ af 

Gets aaa 
xt+y=9 a 68 
eG he coe ch a 

22+ 3y=7 xy =16 
v—y=9 24. 2 +7? =85 


x—y=l 


25. 


26. 


27. 


28. 


29. 


30. 


31. 


32. 


33. 


34. 


35. 


36. 


SIMULTANEOUS QUADRATIC EQUATIONS. 2838 


| 
z—y= 1 
age 
x+t+y= 5 
io. 
zty=8 

1 ae 
ie 5 alll 

ae + y* = 126 


ees 5) 


2 —y> = 56 


ee 28t 


e+ Qey = 24 


2ey +47 = oi 


47 +5xy=14 
Tay+9y? =50 


37. 


39. 


41. 


42. 


44, 


46. 


47. 


“e+ay+y = 39 } 
27+ 32y+y? = 63 


. ae 


ay +- 27" = 40 


Tee ee 
sy +y'=14 


» &+ay+ 27’? = 44 ; 


22° —8ay+2y=16 


#+3y7=31 ; 
4ay+y' = 33 

827+ Try = 82 
w+ 5ay+9y? = 279 


: dae at 
x+ty= 5 
le a 
zty= 3 

: AU wea 
z2—y= 1 
Ce Bod 
e+ty= 1 
ek are 
“e—y= 2 

d eae Cant 
a+y =xy9—1 

: See ed 
z—y =2 


284 SCHOOL ALGEBRA. 


50. BR a 61. . 

x+y =24 x—Yy= 
51. BE Se é. 

8 ay + y = 63 62. = 4¥=1) 
52. 2+ ayt+y= 61 inte 

4 28 A Eas ~ + =e 

x + ay? + y* = 1281 a he 











53. 2 — ay + Y= 4 63. 2? =ax-+ by 
e+ xy? + y* = 21 oe 
tty ,«2—-y_10 30 ples 
54, —— => 64. 2 =2(¢7+0? 
byt eee 3 ee a My 
a? +? = 20 
4 bt 
65 2 2. a+ 
“5: fmm ites tae Oy ae 
t+y «£—y, 9d 
ry = 
8a +4y = 36 
66. P+ Y¥+e+y= 66. gaya 
xy +16 =0 paar 
Mi?) 
57. xz—y—3=0 ; 7~ (a+b) 
2 (2? — y’) = 8xy 
67. xv? —2y ae 
58. atyat sy— y=2ab— 20 
1 4 EC ess 68. yee 
eee ye OG ay = 6 


59. x+y! = 272 i 69. "aT 
sy=v0—40° 


60. eh Sa at 70. 2 a 
z+y =22y-1 x+ty=a+b 


SIMULTANEOUS QUADRATIC EQUATIONS. 285 


Exercise 102. 


1. The area of a rectangle is 60 square feet, and its 
perimeter is 34 feet. Find the length and breadth of the 
rectangle. 


2. The area of a rectangle is 108 square feet. If the 
length and breadth of the rectangle are each increased by 
3 feet, the area will be 180 square feet. Find the length 
and breadth of the rectangle. 


3. If the length and breadth of a rectangular plot are 
each increased by 10 feet, the area will be increased by 400 
square feet. But if the length and breadth are each dimin- 
ished by 5 feet, the area will be 75 square feet. Find the 
length and breadth of the plot. 


4. The area of a rectangle is 168 square feet, and the 
length of its diagonal is 25 feet. Find the length and 
breadth of the rectangle. 


5. The diagonal of a rectangle is 25 inches. If the . 
rectangle were 4 inches shorter and 8 inches wider, the 
diagonal would still be 25 inches. Find the area of 
the rectangle. 


6. A rectangular field, containing 180 square rods, is 
surrounded by a road 1 rod wide. The area of the road 
is 58 square rods. Find the dimensions of the field. 


7. Two square gardens have a total surface of 2137 
square yards. A rectangular piece of land whose dimen- 
sions are respectively equal to the sides of the two squares, 
will have 1093 square yards less than the two gardens 
united. What are the sides of the two squares? 


8. The sum of two numbers is 22, and the difference 
of their squares is 44. Find the numbers. 


286 SCHOOL ALGEBRA. 


9. The difference of two numbers is 6, and their 
product exceeds their sum by 39. Find the numbers. 


10. The sum of two numbers is equal to the difference 
of their squares, and the product of the numbers exceeds 
twice their sum by 2. Find the numbers. ; 


11. The sum of two numbers is 20, and the sum of their 
cubes is 2060. Find the numbers. 


12. The difference of two numbers is 5, and the differ- 
ence of their cubes exceeds the difference of their squares 
by 1290. Find the numbers. 


13. A number is formed of two digits. The sum of the 
squares of the digits is 58. If twelve times the units’ 
digit be subtracted from the number, the order of the 
. digits will be reversed. Find the number. 


14. A number is formed of three digits, the third digit 
being twice the sum of the other two. The first digit plus 
the product of the other two digits is 25. If 180 be added 
* to the number, the order of the first and second digits will 
be reversed. Find the number. 


15. There are two numbers formed of the same two 
digits in reverse orders. The sum of the numbers is 33 
times the difference of the two digits, and the difference of 
the squares of the numbers is 4752. Find the numbers. 


16. The sum of the numerator and denominator of a cer- 
tain fraction is 5; and if the numerator and denominator © 
be each increased by 3, the value of the fraction will be 
increased by 4. Find the fraction. 


17. The fore wheel of a carriage turns in a mile 132 
times more than the hind wheel; but if the circumferences 
were each increased by 2 feet, it would turn only 88 times 
more. Find the circumference of each. 


CHAPTER XX. 
PROPERTIES OF QUADRATICS. 


298. Every affected quadratic can be reduced to the form 
ax’ + bx + e¢=0, of which the two roots are 


_b ,Vbi=4a0 4g 8 vei—4a0 g ogs, 
2a 2a 2a - 2a 


CHARACTER OF THE Roots. 


299. As regards the character of the two roots, there are 
three cases to be distinguished. 


I. If b?’—4ac is positive and not zero, In this case the 
roots are real and unequal. The roots are real, since the ° 
square root of a positive number can be found exactly or 
approximately. If b*—4ac is a perfect square, the roots 
are rational; if 6?—4ae is not a perfect square, the roots 
are surds. 

The roots are unequal, since Vb? — 4.ac is not zero. 


ives pb — 426 is zer0. In this case the two roots are 


~ real and equal, since they both become on 
a 


III. If b’—4ac is negative. In this case the roots are 
wmaginary, since they both involve the square root of a 
negative number. 

The two imaginary roots of a quadratic cannot be equal, 
since 6?—4ac is not zero. They have, however, the same 


~~ 


288 SCHOOL ALGEBRA. 


real part, —2, and the same imaginary parts, but with 
a 


opposite signs; such expressions are called conjugate im- 
aginaries. The expression 6’—4ac is called the discriminant 
of the expression az? -+ bx-+e. 


300. The above cases may also be distinguished as follows: 


CasE I. 0?—4ac>0, roots real and unequal. 
Case II. 6?’—4ac=0, roots real and equal. 
Case III. 0?—4ac <0, roots imaginary. 


301. By calculating the value of 6?—4ae we can deter- 
mine the character of the roots of a given equation without 
solving the equation. 


(1) a? —52+6=0. 


- Here a=1, b=—5, c=6. 
b?—4ac = 25 — 24=1., 


The roots are real and unequal, and rational. 


(2) 827+ 7x2-—1=0. 


Here a=3, d=7, c=—1. 
b?~4ac=49+12=61. 


The roots are real and unequal, and are both surds. 


(3) 427—122+9=0. 


Here a=4 b=-12, c=—¥9. 
b?—4ac= 144-—144=0. 


The roots are real and equal. 


(4) 2a? 382+4=0. 


Here a=2, b=—3, c=4. 
b?—4ac=9 — 32 = — 23, 


The roots are both imaginary. 


\ 


PROPERTIES OF QUADRATICS. 289 

(5) Find the values of m for which the following equa- 

tion has its two roots equal : 
2mx’ + (5m + 2)x-+(4m+1)=0. 
Here a=2m, b=5m+2, c=4m+1. 
If the roots are to be equal, we must have 
b?—4ac=0, or (5m + 2)? —8m(4m+1)=0. 
This gives m= 2, or — =. 
For these values of m the equation becomes 


40? +1224+9=0, and 42?— 42+1=0, 


each of which has its roots equal. 


Exercise 103. 


Determine without solving the character of the roots of 


each of the following equations: . 
Poe = pr -6=0. 6. 627— Tx—3=0. 
Bee + 24 —15=0. 7, 52°7—52—3=0. 
S$. ¢+227+38=0. 8. 2e7—27+5=0. 
4. 32°+7T2#+2=0. 9. 627 +2—7T7=0. 
FO Be ts 1280. 10. bat +82 += 0. 


Determine the values of m for which the two roots of 
each of the following equations are equal: 


11. (m+1)2?+ (m—1)2+m+1=0. 
12. (2m—3)2?+ mz+m—1=0. 

13. 2m2’?+a°?+4274+ 2mex+2m—4=0. 
14. 2m2z?+ 8mx—6=82— 2m — 2’. 
15. m2’?+9x—10=38me—227+2m., 


290 SCHOOL ALGEBRA. 


RELATIONS OF Roots AND COEFFICIENTS. 


302. Consider the equation z7—10z%-+24=0. Resolve 
into factors, (« —6)(a—4)=0. The two values of 2 are 
6 and 4; their sum is 10, the coefficient of 2 with its sign 
changed; their product is 24, the third term. 


308. In general, representing the roots of the quadratic 
equation aa’?+ b4+c¢=0 vt 7, and 7,, we have (§ 285), 


5 Vb? —4a0¢ 4 a0 
Bik ay o— ene 
‘ Mie cianee'| Vo? —4a¢ 4ac 
: 2a 2a 
: b 
Adding, TY) + LE areata peek 
a 
Rite: Cc 
multiplying, Tits oe 


If we divide the equation ax’ -+ bx+c=0 through by 
; b ; 
a, we have the equation 2 +o2+—=0; this may be 
written z’?+ px-+gq=0 where p =< g =<. 

It appears, then, that if any quadratic equation be made 
to assume the form 2’-++ px + ¢=0, the following relations 
hold between the coefficients and roots of the equation : 

(1) The sum of the two roots is equal to the coefficient 
of x with its sign changed. 

(2) The product of the two roots is equal to the constant 
term. 

Thus, the sum of the two roots of the equation 


v’—Tx+8=0 
is 7, and the product of the roots 8. 


PROPERTIES OF QUADRATICS. 291 


304, Resolution into Factors. By § 303, if 7, and 7 are 
the roots of the equation 2+ px+q=0, the equation 


may be written 
2—(n+r)e+rnr,= 0. 


The left member is the product of x —7, and x — 7, so 
that the equation may be also written 


(2 —1,)(x#— 1.) = 0. 


It appears, then, that the factors of the quadratic expres- 
sion x? + px-+q are x—7, and x— 7, where 7, and r, are 

the roots of the quadratic equation 2° + px+q=0. 

- he factors are real and different, real and alike, or 
imaginary, according as 7, and r, are real and unequal, 
real and equal, or imaginary. 

If 7,=7;, the equation becomes (x — 17) (2 — 7) = 0, or 
(c—7,)?=0; if, then, the two roots of a quadratic equa- 
tion are equal, the left member, when all the terms are 
transposed to that member, will be a perfect square as 
regards x. 


305. If the equation is in the form az’?+ bx + ¢=0, the 
left member may be written 


(v b C 
a\ x +045), 
a(e— 7) (% — 72). 


306. If the roots of a quadratic equation are given, we 
can readily form the equation. 


Ex. Form the equation of which the roots are 3 and — 2. 
The equation is (x#—3) (« + 5) = 0, 


or (a — 3) (2% + 5) =0, 
or 207 —x2—15=0, 


992, SCHOOL ALGEBRA. 


3807. Any quadratic expression may be resolved into 
factors by putting the expression equal to zero, and solving 
the equation thus formed. 


(1) Resolve into two factors 7? —5x +3. 


Write the equation 
v?—5xe4+3=0., 
Solve this equation, and the roots are found to be 


5 + nae Bee 5 vis 


Therefore, the factors of a? —5a2+3 are 
ej Dt Vib ng eciD iene 
2 2 
(2) Resolve into factors 32°—4z-+ 5. 


Write the equation 
3a7—42+5=0. 
Solve this equation, and the roots are found to be 


2b Vet aS ee ed 
rir ee eee 
Therefore, the expression 327—4a+5 may be written (¢ 305), 
3 (2-24¥%= 3) (ee a 
t 3 3 


Exercise 104. 


Form the equations of which the roots are 


187; 6. opie Sea 9. 84+ V2,3— v2. 
2. 5,—8. 6. —1,,—1]%. (10. 1+ Vee 
3. 14,—2. 7. 18, —44. 11. a,a—b. 
4. 4, 21. 8. - = 12. at+b,a—b. 
Resolve into factors, real or imaginary : 
13. 122? +2—1. 16. v2 —22+8. 
14. 827— 142 — 24, 17. 2te+l. 


15. 2? —2Q2 —2. 18. 2?—22+9. 


CHAPTER XXII. 
RATIO, PROPORTION, AND VARIATION. 


308. Ratio of Numbers. The relative magnitude of two 
numbers is called their ratio when expressed by the indi- 
cated quotient of the first by the second. Thus the ratio 


of a to 6 is 5 or a+b, or a:b. 


The first term of a ratio is called the antecedent, and the 
second term the consequent. When the antecedent is equal 
to the consequent, the ratio is called a ratio of equality; 
when the antecedent is greater than the consequent, the 
ratio is called a ratio of greater inequality; when less, a ratio 
of less inequality. 

When the antecedent and consequent are interchanged, 
the resulting ratio is called the inverse of the given ratio. 
Thus, the ratio 3:6 is the znverse of the ratio 6: 8. 


809. A ratio will not be altered if both its terms are 
multiplied by the same number. For the ratio a:6 is 


represented by 5 the ratio ma: mb is represented by 1 
m 
and since “@=" we have ma:mb=a:b. 
mb 6b 


310. A ratio will be altered if different multiphers of its 
terms are taken; and will be increased or diminished accord- 
ing as the multiplier of the antecedent is greater than or 
less than that of the consequent. Thus, 


294 SCHOOL ALGEBRA. 


If m>n, If m<n, 
then ma>na, then ma< na, 
d lake Seo alge d ha ae gilt 
me nb a nb’ by nb “nb 
but (cn LS but deans NS 
- nb b ‘ nbd 
Ma & . bbs ie Lo 
Pas * RE re: 
or ma:nb>a:b. or ma:nb<a:b. 


311. Ratios are compounded by taking the product of the 
fractions that represent them. Thus, the ratio compounded 
of a:b and ¢: dis ac: bd. 

The ratio compounded of a:6 and a:6 is the duplicate 
ratio a’: 6"; the ratio compounded of a:b, a: 6, and a:6 is 
the triplicate ratio a*: 6°; and so on. 


312. Ratios are compared by comparing the fractions 
that represent them. 


Thus, QO Ssor 7 6a. 
according as a or < : 
as a> or << 
as ad > or < be. 


313. Proportion of Numbers. Four numbers, a, 8, ec, d, are 
said to be in proportion when the ratio a: 6 is equal to the 
ratio ce: d. 

We then write a:b =c:d, and read this, the ratio of a to 
6 equals the ratio of ¢ to d, or ais to 8 as ¢ is to d. 

A proportion is also written a:6::e:d. 

The four numbers, a, 0, c, d, are called proportionals; a 
and d are called the extremes, 6 and ¢ the means, 


RATIO AND PROPORTION. 295 


314, When four numbers are in proportion, the product 
of the extremes is equal to the product of the means. 


For, if arb=c% d, 
th ME 
en F, 


Multiplying by dd, ad=be. 


The equation ad=be gives a=" p= so that an 
extreme may be found by dividing the product of the 
means by the other extreme; and a mean may be found by 
dividing the product of the extremes by the other mean. 
If three terms of a proportion are given, it appears from 
the above that the fourth term can have one, and but one, - 
value. 


315, If the product of two numbers is equal to the prod- 
uct of two others, either two may be made the extremes of 
a proportion and the other two the means. 


For, if ad = be, 


fae ad_ be 
then, dividing by dd, bd bd 
or pa. 
std, 
Res eB ea ee A 


316. Transformations of a Proportion. If four numbers, a, 
b, c, d, are in proportion, they will be in proportion by: 


I, Inversion; that is, 6 will be to a as dis to ec. 
For, if Ob == oS.d, 
then 


296 SCHOOL ALGEBRA. 








and neta iste 
bad, 
or ee SE 
Cac 
21D soli ah es 
II, Composition; that is, a+ will be to 6 as e+d is to d. 
For, if Tied ak cielo 
h egos 
then i a 
and Fe heros i 
ee ater 


-atb:b=c+d:d. 
III. Division; that is, a—6 will be to 6 as e— dis to d. 








For, if Tae tow tes 2 
then Fee 
and lees 
Re GD. saa 

b d 


” a—b:b=c—d:d.. 


IV. Oomposition and Division; that is, a+ will be to 
a—base+dis to c—d. 














For, from II., ate_ctd 
and from III., oe 
a+b etd 








Dividing, 5 
a c— 
*atb:a—b=ce+ 


RATIO AND PROPORTION. 297 


V. Alternation; that is, a will be to c as 6 is to d. 


For, if a b= e7'd. 
then ae 
war. b 6b be 
Multipl pees rl Ee a 
ultiplying by : ae a 
ae ao 
et 70 

eet ees D4. 


317. In a series of equal ratios, the sum of the antecedents 


is to the sum of the consequents as any antecedent is to its 
consequent. 


Orsay aime dow 
y may be put for each of these ratios 
Then C=yr, ee Cy Lay, 


a= OT. ONE = FTR g = AP. 

“ atetetg=(64+d+ft+h)r. 

Beri ener Fi ae 
b+d+tfth b 

“ atetetg:b+d+f+th=a: 6, 


In like manner it may be shown that 
ma+ne-+petgqg:mb+nd+pft+qh=a:b. 
318, Continued Proportion. Numbers are said to be in 


continued proportion when the first is to the second as the 
second is to the third, and so on. Thus, a, 8, ¢, d, are in 


continued proportion when 


298 SCHOOL ALGEBRA. 


319. Ifa, 6, ¢ are proportionals, so that a:b =b:c, then 
6 is called a mean proportional between a@ and c, and ¢ is 
called a third proportional to a and 6. 


If a:b6=b:c, then b= Vac. 
For, if i SaID Se 
a_t 
then ae 
and Of Sac: 
b= Vae. 


320. The products of the corresponding terms of two or 
more proportions are in proportion. 


For, if a: b= ou 


then 


Taking the product of the left members, and also of the 
right members of these equations, 


. ack: bfl = cgm: dhn. 


$21, Like powers, or like roots, of the terms of a propor- 
tion are in proportion. 


For, if Wit DiS orate 
then ; == e 
Raising both sides to the nth power, 
a” aoe a 
ai ot Me 


RATIO AND PROPORTION. 299 


322, The laws that have been established for ratios 
should be remembered when ratios are expressed in frac- 
tional form. 

2 2 Aye ar y 

Gece: Sette eee 

V’—x-—-1l #+ne-2 
By composition and division, 
Sree MS Serna 
B(x +1). — 2(2 —2) 

This equation is satisfied when a=0. For any other value of z, 

we may divide by 2’. 


We then have l ! 








and therefore 


(2) If a:b =c:d, show that 
v+ab:0?—ab=c’+ cd: d’?—cd. 














a ce 
: see 
a+b c+d 
sales Gm bi eed 
dk Bas ae 
and Deng 
a at+b_¢ etd. 
i <0) ee Gee 
that is Stree acne 





ab @—cd 
or a+ab:6?—ab=c?+ cd: d*?—cd. 


300 SCHOOL ALGEBRA. 


(3) If a:b =c:d, and a is the greatest term, show that 
a+d is greater than b+ c. 








Since as a and. a& > ¢, (1) 
the denominator b> d. 
From (1), by division, © $ aes = se (2) 
Since psa 
from (2), a—b>c—d, 
Now, b+d=b+d, 
Adding, a+d>b+e. 


323. Ratio of Quantities. To measure a quantity of any 
kind is to find out how many times it contains another 
known quantity of the same kind, called the unit of measure. 
Thus, if a line contains 5 times the linear unit of measure, 


one yard, the length of the line is 5 yards. 


324, Commensurable Quantities. If two quantities of the 
same kind are so related that a unit of measure can be 
found which is contained in each of the quantities an in- 
tegral number of times, this unit of measure is a common 
measure of the two quantities, and the two quantities are 
said to be commensurable. 

If two commensurable quantities are measured by the 
same unit, their ratio is simply the ratio of the two numbers 
by which the quantities are expressed. Thus, + of a foot is 
a common measure of 24 feet and 32 feet, being contained 
in the first 15 times and in the second 22 times. 

The ratio of 24 feet to 32 feet is therefore the ratio of 
15: 22. 

Evidently two quantities different in kind can have no 
ratio, 


4 


RATIO AND PROPORTION. 301 


325, Incommensurable Quantities. The ratio of two quan- 
tities of the same kind cannot always be expressed by the 
ratio of two whole numbers. Thus, the side and diagonal 
of a square have no common measure; for, if the side is a 
inches long, the diagonal will be a2 inches long, and no 
measure can be found which will be contained in each an 
integral number of times. 

Again, the diameter and circumference of a circle 
have no common measure, and are therefore incommen- 
surable. 

In this case, as there is no common measure of the two 
quantities, we cannot find their ratio by the method of 
§ 324. We therefore proceed as follows: 


Suppose a and 6 to be two incommensurable quantities 
of the same kind. Divide 6 into any integral number (7) 
of equal parts, and suppose one of these parts 1s contained in 
a@ more than ™ times and less than m-+1 times. Then the 
ratio ; sa —, but < LOnrae that is, the value of : les 

n 


between — an 


m q mel, 
n 


The error, therefore, in taking either of these values for 


@ is less than 4. But by increasing 7 indefinitely, 1 can 
n n 


be made to decrease indefinitely, and to become less than 
any assigned value, however small, though it cannot be 
made absolutely equal to zero. 

Hence, the ratio of two incommensurable quantities can- 
not be expressed exact/y in figures, but it may be expressed 
approxvmately to any desired degree of accuracy. 

Thus, if 6 represent the side of a square, and a the 


diagonal, 
ie 


302 SCHOOL ALGEBRA. 


Now V2 = 1.41421856.....,a value greater than 1.414218, 
but less than 1.414214. . 

If, then, a mzdlionth part of 6 is taken as the unit, the 
1414213 .. , 1414214 
1000000 1000000" 
and therefore differs from either of these fractions by less 
than bani 

1000000 

By carrying the decimal farther, a fraction may be found 
that will differ from the true value of the ratio by less than | 
a billionth, a trillionth, or any other assigned value what- 
ever. 








value of the ratio ;, lies between 


826. The ratio of two incommensurable quantities is an 
incommensurable ratio, and is a fixed value toward which 
its successive approximate values constantly tend as the 
error 1s made less and less. 


827. Proportion of Quantities. In order for four quanti- 
ties, A, B, C, D, to be in proportion, A and B must be of 
the same kind, and C' and D of the same kind (but Cand 
D need not necessarily be of the same kind as A and 8B), 
and in addition the ratio of A to B must be equal to the 
ratio of C to D. 

If this be true, we have the proportion 


AA Bata gy: 


When four quantities are in proportion, the numbers by 
which they are expressed are four abstract numbers in 
proportion. 


828, The laws of § 816, which apply to proportion of 
_numbers, apply also to proportion of quantities, except that 
alternation will apply only when the four quantities in 
proportion are a// of the same kind. 


RATIO AND PROPORTION. 303 


Exercise 105. 


1. Find the duplicate of the ratio 3° 4. 

2. Find the ratio compounded of the ratios 2: 3, 3: 4, 
S714: 8. 

3. Find a third proportional to 21 and 28. 

4. Find a mean proportional between 6 and 24. 

5. Find a fourth proportional to 3, 5, and 42. 

6. Find #if 5++-2:1l1—27=8:5. 

7. Find the number which must be added to both the 


terms of the ratio 3:5 in order that the resulting ratio 
may be equal to the ratio 15: 16. 


If a:b =c: d, show that 


me od — ¢*: dd’. 10. @—OB:?-aP=a':e. 
eri eo. == Cr °C", 11. 2a+6:2c+d=0d:d. 
. da—b:5ce—d=a:e. 

. a—8b:a+38b=c—8d:c+8d. 

~ @+ab+:a—abtPHe+ed+a’?:e—ced+d’. 


Find # in the proportion 


15. 
16. 
19. 


45°68 = 90* 2. 17. x: l4A=—13: 14. 
pag a: 1, IBA LA. 


Find two numbers in the ratio 2:3, the sum of whose 


squares is 325. 


20. 


Find two numbers in the ratio 5:8, the difference 


of whose squares is 400. 


21. 


Find three numbers which are to each other as 


2:3:5, such that half the sum of the greatest and least 
exceeds the other by 25. 


304. SCHOOL ALGEBRA. 


22. Find cif 6%#—a:4r7—b=—82+06: 2z-a. 

23. Find x and y from the proportionals 
er:y=ue+y:42; ri y=e— yi. 

24. Find x and y from the proportionals 
2ety:y=syi2y—az; 
2¢2+1:24+6=y:y+2. 

25 dt Go OAR Ed a Dee how that 


=e 


atb—e—d a—b—cta 


et 
De a 


VARIATION. 


329, A quantity which in any particular problem has a 
fixed value is called a constant quantity, or simply a constant ; 
a quantity which may change its value is called a variable 
quantity, or simply a variable, 

Variable numbers, ike unknown numbers, are generally 
represented by a, y, z, etc.; constant numbers, like known 
numbers, by a, 8, e, etc. 


830. Two variables may be so related that when a value 
of one is given, the corresponding value of the other can be 
found. In this case one variable is said to be a function 
of the other; that is, one variable depends upon the other 
for its value. Thus, if the rate at which a man walks is 
known, the distance he walks can be found when the time 
is given; the distance is in this case a function of the time. 


331, There is an unlimited number of ways in which 
two variables may be related. We shall consider in this 
chapter only a few of these ways. 


332, When x and y are so related that their ratio is 
constant, y is said to vary as x; this is abbreviated thus: 


VARIATION. 305 


you. The sign o, called the sign of variation, is read 
“varies as.’ Thus, the area of a triangle with a given 
base varies as its altitude; for, if the altitude is changed 
in any ratio, the area will be changed in the same ratio. 

In this case, if we represent the constant ratio by m, 


y 
Yi tZ=M, or = M; “Y= me. 


Again, if y’, z' and y", 2" be two sets of corresponding 
values of y and 2, then 


yt ah yl al" 
or ey ae a § 316, V. 
1 


z 
is constant, y is said to vary wversely as x; this is written 


1 


ya -. Thus, the time required to do a certain amount of 
xe 


333, When x and yare so related that the ratio of y to 


work varies inversely as the number of workmen employed ; 
for, if the number of workmen be doubled, halved, or 
changed in any ratio, the time required will be halved, 
doubled, or changed in the inverse ratio. 

In this case, y =m; ox: y=—, and xy=m,; that is, 


the product xy is constant. 


i 1 
As before, rae ap ay 
hay a ally"! 
or, ED Bh ae Ga $315 


334, If the ratio of y: xz is constant, then y is said to 
vary jointly as x and z. 
In this case, y= mzz, 


and he eels celal 


306 SCHOOL ALGEBRA. 


835. If the ratio y:~ is constant, then y varies directly 
z 
as x and wmversely as z. 


In this case, ea, 
and Je 


336. Theorems. 
I. If you, and xz, then y oz. 
For y=me«z and «—nz, 
i 
“. Y Varies as 2. 
Il. If you, and xz, then (y+z2)oz. 
For y= mz ond z—nz. 
 YRZ=(MAN)EZ; 
“. Y +2 varies as x. 
Ill. If yox when z is constant, and yoz when @ is 
constant, then y « #z when z and z are both variable. 
Let 2!, y', z', and x", y", 2" be two sets of corresponding 
values of the variables. 


Let # change from 2! to x", z remaining constant, and let 
the corresponding value of y be Y. 


Then IPAS Gran AME nF (1) 
Now let z change from z! to 2", x remaining constant. 
Then Y: ay! — gle Zl (2) 
From (1) and (2), 
WV y"V = a'a! vale", § 320 
or Ue aft ag gh lake ; 
or of ele! ol bat aE § 816, V. 


. Y e . 
.“. the ratio pe constant, and y varies as wz. 


VARIATION. 307 


In like manner it may be shown that if y varies as each 
of any number of quantities 2, z, wu, etc., when the rest are 
unchanged, then when they all change, y « xzu, etc. Thus, 
the area of a rectangle varies as the base when the altitude 
is constant, and as the altitude when the base is constant, 
but as the product of the base and altitude when both vary. 


337, Examples. 


(1) If y varies inversely as x, and when y= 2 the cor- 
responding value of x is 36, find the corresponding value 
of when y= 9. 

Here y=", or m=ay. 

wv 
.m=2X36= 72. 


If 9 and 72 be substituted for y and m respectively in 


_m 
x 
9 
the result is 9= Lie or 9a = 72. 
x 
*a2=8. Ans. 


(2) The weight of a sphere of given material varies as 
its volume, and its volume varies as the cube of its diam- 
eter. Ifa sphere 4 inches in diameter weighs 20 pounds, 
find the weight of a sphere 5 inches in diameter. 


Let W represent the weight, 
V represent the volume, 
D represent the diameter. 
Then Wee V and Vee D3, 
SAG Vy we oe @ 336, I. 
Put W=mD>; 
then, since. 20 and 4 are corresponding values of W and D, 
20 = m X 64. 
20 5 


ie W= aan Dd. 
*. when D=5, W= 5; of 125 = 3974. 


308 SCHOOL ALGEBRA. 


Exercise 106. 
1. Ifxecy,and if y=3 when x=5, find # when yis 5. 


2. If W varies inversely as P, and W is 4 when Pis 
15, find W when P is 12. . 


8. If xocyand yz, show that we 0 y?, 


4. If rot and oe show that x « z. 
y zZ 


5. If # varies inversely as 7?—1, and is equal to 24 
when y= 10, find when y=5. 


6. If x varies as ee and is equal to 3 when y=1 
Tet S 


and z= 2, show that zyz= 2(y-+2). 


. . 1 . 
7. If x—y varies inversely as bis and #-+y varies 

1 
inversely ‘as uae find the relation between a and z if 


2=1, y=38, when 25. 


8. The area of a circle varies as the square of its radius, 
and the area of a circle whose radius is 1 foot is 3.1416 
square feet. Find the area of a circle whose radius is 20 
feet. 


9. The volume of a sphere varies as the cube of its 
radius, and the volume of a sphere whose radius is 1 foot is 
4.188 cubic feet. Find the volume of a sphere whose radius 
is 2 feet. 


10. If a sphere of given material 3 inches in diameter 
weighs 24 lbs., how much will a sphere of the same material 
weigh if its diameter is 5 inches? 


VARIATION, | 309 


11. The velocity of a falling body varies as the time 
during which it has fallen from rest. If the velocity of a 
falling body at the end of 2 seconds is 64 feet, what is its 
velocity at the end of 8 seconds? 


12. The distance a body falls from rest varies as the 
square of the time it is falling. If a body falls through 144 
feet in 8 seconds, how far will it fall in 5 seconds? 


The volume of a right circular cone varies jointly as 
its height and the square of the radius of its base. 

If the volume of a cone 7 feet high with a base whose 
radius is 8 feet is 66 cubic feet : 


13. Compare the volume of two cones, one of which is 
twice as high as the other, but with one half its diameter. 


14. Find the volume of a cone 9 feet high with a base 
whose radius is 8 feet. 


15. Find the volume of a cone 7 feet high with a base 
whose radius is 4 feet. 


16. Find the volume of a cone 9 feet high with a base 
whose radius is 4 feet. 


17. The volume of a sphere varies as the cube of its 
radius. If the volume is 1792 cubic feet when the radius 
is 34 feet, find the volume when the radius is 1 foot 6 
inches. 


18. Find the radius of a sphere whose volume is the sum 
of the volumes of two spheres with radii 34 feet and 6 feet 
respectively. 


19. The distance of the offing at sea varies as the square 
root of the height of the eye above the sea-level, and the 
distance is 8 miles when the height is 6 feet. Find the 
distance when the height is 24 feet. 


CHAPTER XXII. 


PROGRESSIONS. 


338, A succession of numbers that proceed according to 
some fixed law is called a series; the successive numbers are 
called the terms of the series. 

A series that ends at some particular term is a finite series; 
a series that continues without end is an infinite series. 


339, The number of different forms of series is unlimited; 
in this chapter we shall consider only Arithmetical Series, 
Geometrical Series, and Harmonical Series. 


ARITHMETICAL PROGRESSION. 


340, A series is called an arithmetical series or an arith- 
metical progression when each succeeding term is obtained 
by adding to the preceding term a constant difference. 

The general representative of such a series will be 

a, at+d,at+2d, a+8d..... 
in which a is the first term and d the common difference ; 


the series will be increasing or decreasing according as d is 
positive or negative. 


341. The nth Term. Since each succeeding term of the 
series 18 obtained by adding d to the preceding term, the 
coefficient of d will always be one less than the number of 
the term, so that the nth term is a+(n—1)d. 

If the nth term be represented by 7, we have 


l=a+(n—1)d. 1. 


ARITHMETICAL PROGRESSION. olL 


842, Sum of the Series. If Z denotes the nth term, a the 
first term, 2 the number of terms, d the common difference, 
and s the sum of n terms, it is evident that 


s= a +(atd)+(a4+2d)+---+(U—d)+ 7, or 
s= 1 4(0—-d)t+ (2d) tt td)+ 

“. 2s=(a+l)+(a+l) +(atd) +e +(atl) +(a4+0) 
=n(a-+ 0). 


n 


s=5(a+ 4). II. 


343, F'rom the equations 
l=a+(m—1)d, i 
s=5 (a+!) if. 


any two of the five numbers a, d, /, n, s, may be found when 
the other three are given. 


Here | ‘G=2, d=3 
From I., [= 2 


Substituting in II., $= 


(2) The first term of an arithmetical series is 3, the last 
term 31, and the sum of the series 186. Find the series. 


_ From I., 31=34(n—1)d. (1) 
From II., 136 = 530 +81). (2) 
From (2), n= 8, 


Substituting in (1), d=4., 
miueeseries. ig oO, 1, 11, 16; 19, 23, 27, 31. 


SY SCHOOL ALGEBRA. 


(3) How many terms of the series 5, 9, 18, ....., must be 
taken in order that their sum may be 275? 


From L,, 1=5+(n—1)4. 
v l=4n41. (1) 
From II., 275 = rAG + 1). (2) 


Substituting in (2) the value of 7 found in (1), 
275 = 5 (tn +6), 
or 2n? + 3n= 278. 


This is a quadratic with n for the unknown number. 


Complete the square, 
16 n? + () + 9 = 2209. 
Extract the root, 4n+3=+ 47. 


Therefore, n=11, or — 121. 


We use only the positive result. 


(4) Find x when d, J, s are given. 


From L., a=l1—(n—1)d. 
From IL., a= 28s — In. 
n 
2s—In 
Therefore, l—(n —1)d==—. 
n 


“. In—dn? + dn=2s—In, 
“. dn? — (214+ d)n=— 2s. 
This is a quadratic with n for the unknown number. 
Complete the square, 
4d°n? —() + (21+ dl? =(21+ d)— 8ds. 
Extract the root, 
2dn —(21+d)=+ V(21+ dy — 8ds. 


gl+d+V(21+ d)—8ds 
2d 


Therefore, n= 


ARITHMETICAL PROGRESSION. 313 
(5) Find the series in which the 15th term is — 25 and 
Alst term — 41. 


If a is the first term and d the common difference, the 15th term 
is a + 14d, and the 41st term is a + 40d. 


Therefore, a+14d=— 25, (1) 
and a+40d=—41.- (2) 
Subtracting, —26d= 16. 
ben ee) 
eis 
Substituting in (1), a = — 16,5. 


Hence, the series is — 1635,, — 17, — 173%, ..... 


344, The arithmetical mean between two numbers is the 
number which stands between them, and makes with them 
an arithmetical series. 

If a and 6 represent two numbers, and A their arithmet- 
ical mean, then a, A, 6 are in arithmetical progression, and 
by the definition of an arithmetical series, § 340, 


A—a=d, 

and 6—A=d. 
. A-—a=b—-A 
até 
AES omen 


345, Sometimes it is required to insert several arithmeti- 
cal means between two numbers. 


Ex. Insert six arithmetical means between 3 and 17. 
Here the whole number of terms is eight; 3 is the first term and 
17 the eighth. 
By L,, ee 
d= 2. 
The series is 3, [5, 7, 9, 11, 18, 15,] 17, the terms in — 
brackets being the means required, 


314 SCHOOL ALGEBRA. 


346, When the sum of a number of terms in arithmetical 
progression is given, it is convenient to represent the terms 
as follows: 


Three terms by x—y, x, x+y; 
four terms by t+ SY, BY, oY eee 


and so on. 


Ex. The sum of three numbers in arithmetical progres- 
sion is 36, and the square of the mean exceeds the product 
of the two extremes by 49. Find the numbers. 


Let «—y, x, x+y represent the numbers. 
Then, adding, 3.2 = 36. 
*, @ = 12. 
Putting for w its value, the numbers are 
12—y, 12, 1244, 
Then (12)?—(12—y)(12 + y) = the excess. 


But 49 = the excess. 
Therefore, 144 — 144 + y? = 49. 
a ee ie 


The numbers are 5, 12, 19; or 19, 12, 5. 


Exercise 107. 
l=a+(n—1)d; s= (a +1) =5(2a+(n—1)d} 


Find / and s, if 


1 a=l, d=4, n=18. -4.. a= 63. ee 
Bf cei) 2 Onan eens a=%, d=2, oe 
3. eee d=6,n=30,"° 6. a@=3n, d= 2 


9 


ARITHMETICAL PROGRESSION. 315 


Find d and sg, if 
pence ac lot. te 13. 


Sara — 0) b= 2005 n= 51, 
Sie lous hs 6, I. SA. 
FOr Ose 1450) 2 — 21. 


Insert eight arithmetical means between 


ite io and 76. 13. 1 and 3. 
2 4 
ee GAT 14. 47 and 2. 


Find a and s, if 
oe et = 149 2 22, 16. d= 21, (= 242, n= 12. 


Find n and s, if 
Reet 000, ad=—9. 18.034 7=—10,d=—2. 


Find d and 7, if 
fee a—At, (= 54, s=999. -20.a=—2, 7=—87, s=80l. 


Find d and J, if 
meee. 2—14,5—1050) 22° a—1, n= 20, s= 800. 


Find a and d, if 
free a (,s— 105. 24. /=105,n—16, s— 840. 


Find a and J, if 


25. n=21,d=4,s=1197. 26. n=25, s=—75, d=5 


Find and J, if 
freee ooo. a-—9,d=—8. 28: s— 7198, a= 18; d=6. 


Find a and 1, if 
Some o25 a= 5,1=77. -30. s=1008,d=—4; /=88, 


316 SCHOOL ALGEBRA. 


Find the arithmetical series in which 
31. The 15th term is 25, and the 29th term 46. 


How many terms must be taken of 


32. The series —16, —15, —14, ..... to make — 100? 
33. The series 20, 18%, 174, .....to make 1621? 


34. The sum of three numbers in arithmetical progres- 
sion is 9, and the sum of their squares is 29. Find the 
numbers. 


35. The sum of three numbers in arithmetical progres- 
sion is 12, and their product is 60. Find the numbers. 


GEOMETRICAL PROGRESSION. 


847. A series is called a geometrical series or a geometrical 
progression when each succeeding term is obtained by mul- 
tiplying the preceding term by a constant multipher. 

The general representative of such a series will be 


Oh, OF SI0T OR" Cree : 


in which a is the first term and r the constant multiplier 
or ratio. 

The terms increase or decrease in numerical magnitude 
according as 7 is numerically greater than or numerically 
less than unity. 


848, The nth Term. Since the exponent of 7 increases by 
one for each succeeding term after the first, the exponent 
will always be one less than the number of the term, so 
that the nth term is a7”-}. 


If the nth term is represented by /, we have 


pe ae) ik 


GEOMETRICAL PROGRESSION. 317 


349, Sum of the Series. If / represent the nth term, a the 
first term, n the number of terms, 7 the common ratio, and 
s the sum of 7 terms, then 


s=atartarte- art > mh) 
Multiply by 7, 
rs =ar + ar? + ar ar" + ar". (2) 


Subtracting the first equation from the second, 


T8§ —s = ar" — a, 





or (r —l)s=aG"— 1). 
eee =: Atos); a1. 
r—1 
Since /= ar", r/ = av”, and II. may be written 
rl —a, 
sa Jane 
F y—l 


350, From the two equations I. and II., or the two equa- 
tions I. and III., any two of the five numbers a, 7, J, n, s, 
may be found when the other three are given. 


(1) The first term of a geometrical series is 3, the last 
term 192, and the sum of the series 881. Find the number 
of terms and the ratio. 


From I., 1923) (1) 


From IIL, fale esa (2) 
| eae 
From (2), 7 = 2, 
Substituting in (1), aK BA. 
en =n. 


The series is 8, 6, 12, 24, 48, 96, 192. 


318 SCHOOL ALGEBRA. 


(2) Find Z when 7, 1, s are given. 











From I., pica 
7r-l 

rl — = 

Substituting in IIL., g§= = , 
Yi — 

(r—1)s= usar: 2) 
ere 

ea (r— 1)"; 
rr — | 


351. The geometrical mean between two numbers is the 
number which stands between them, and makes with them 
a geometrical series. 

If a and 6 denote two numbers, and G their geometrical 
mean, then a, G, 6 are in geometrical progression, and by 
the definition of a geometrical series, § 347, | 


G b 
ay and ae 
Wint fered 
rags 
. G=Vab. 


352, Sometimes it is required to insert several geometri- 
cal means between two numbers. 
Ex. Insert three geometrical means between 3 and 48. 


Here the whole number of terms is five; 3 is the first term and 48 
the fifth. 


By 1, 48 = 374, 
rt = 16. 
r=+ 2. 


The series is one of the following: 
Boyt eo eeke: 24), 48; 
3, [-—6, 12, —24,] 48. 
The terms in brackets are the means required. 


GEOMETRICAL PROGRESSION. 319 


3538, Infinite Geometrical Series. When 7 is less than 1, the 
successive terms become numerically smaller and smaller ; 
by taking 7 large enough we can make the nth term, ar"~’, 
as small as we please, although we cannot make it absolutely 


zero. 1 i 
The sum of n terms, ie ba a asl ; 
7 — l—r l- r 











“may be written 








“— by the fraction - ; by taking 


—r 7 
enough terms we can make /, and consequently the fraction 


this sum differs from i 


ha , as small as we please; the greater the number of 
—*7 


a 
terms taken the nearer does their sum approach sare 








Hence is called the sum of an infinite number of 


—/ 
terms of the series. 





(1) Find the sum of the infinite series 1 — ; a i — ‘ Se 
Here, a =1, een 
2 
The sum of the series is l or z Ans 
| ekg ee 


We find for the sum of n terms ; - : ig iin this sum evidently 
v 


2 : 
approaches pias me increasen: 


(2) Find the value of the recurring decimal 
0.12135185 ..... 


Consider first the part that recurs; this may be written 
135 


= +..., and the sum of this series is 100000” 


100000 100000000 qe 1 
mae 1000 
which reduces to 740" Adding 0.12, the part that does not recur, we 


obtain for the value of the decimal 12 a 22 or 449 Ans. 


100 740’ 3700 


320 SCHOOL ALGEBRA. 


Exercise 108. 


a(rm—1)_rl—a 


fae ye ae 
r—1 r—l 


Find Z/ and s, if 

LS yn ey eels 2d raz n=l. 
Find r and s, if 

Sh eh as 4 le 4: 

Qe de sala a0 Os 


ss 


5. Insert 1 geometrical mean between 14 and 686. 

6. Insert 38 geometrical means between 31 and 496. 

%. Find @ and ¢,11.( = 128, 7-2) =e 

8. Find sand n, if a=9, 7= 2304, r=2. 

9. Find r and n, if a=2, 7= 1458, s = 2186. 

10. If the 5th term is 3 and the 7th term 34, find the 
series. 


11. Find three numbers in geometrical progression whose 
sum is 14, and the sum of whose squares 84. 


Sum to infinity : 


bey ay 1 1 ier eed Ae 2 2 new te OEE 1 i 
D4 7 49 4° 16 
1 9 4 
13. Dee LBA Dio eens 17.8, —2 tee 
3 8, ? 9 ? ? 5’ q 5] ? 3 
Find the value of 
18. 0.16. 20. 0.86. 22. 0.736. 


19. 0.378. 21. 0.54. 23. 0.363. 


HARMONICAL PROGRESSION. O21 


HARMONICAL PROGRESSION. 


354, A series is called a harmonical series, or a harmonical 
progression, when the reciprocals of its terms form an arith- 
metical series. 

‘The general representative of such a series will be 

Pee oay MUL eres wines 2D ve 4, 
pera fat ated Oh aly (on 

Questions relating to harmonical series are best solved 
by writing the reciprocals of its terms, and thus forming 
an arithmetical series. 





355. If a and 6 denote two numbers, and # their har- 
monical mean, then, by the definition of a harmonical series, 





plete ste. 
Fe hae cba 
_ 2ab- 
evs, 


356. Sometimes it is required to insert several harmoni- 
cal means between two numbers. 


Ex. Insert three harmonical means between 3 and 18. 


Find the three arithmetical means between 1 and . 








These are found to be = =f mi therefore, the harmonical means 
Pe or 348 Bh, 8: 
19 14 9 
857, Since A aia b and G@= Vab, and H= 2ab 
2 a+b 


wes _ ri CAL 


That is, the geometrical mean between two numbers is 
also the geometrical mean between the arithmetical and 
harmonical means of the numbers. 


4 SCHOOL ALGEBRA. 


Exercise 109. 
1. If a, 6,c¢ are in harmonical progression, show that 
a—b:b-c=a:e. 


2. Show that if the terms of a harmonical series are all 
multiplied by the same number, the products will form a 
harmonical progression. 


3. The second term of a harmonical series is 2, and the 
fourth term 6. Find the series. 


4. Insert the harmonical mean between 2 and 3. 


5. Insert 2 harmonical means between 1 and 


6. Insert 5 harmonical means between 1 and * 


7. The first term of a harmonical progression is 1, and 


the third term ; Find the 8th term. 


8. The first term of a harmonical progression is 1, and 
the sum of the first three terms is 18. Find the series. 


9. Ifa is the arithmetical mean between 6 and e¢, and b 
the geometrical mean between a and c, show that ¢ is the 
harmonical mean between a and 6. 


10. The arithmetical mean between two numbers exceeds 
the harmonical mean by 1, and twice the square of the 
arithmetical mean exceeds the sum of the squares of the 
harmonical and geometrical means by 11. Find the num- 
bers. 


CHAPTER XXIII. 
PROPERTIES OF SERIES. 


358, Convergent and Divergent Series. By performing the 


indicated division, we obtain from the fraction L the 
—2£ 





infinite series 1+ 2+ 2?+2°+...... This series, however, 
is not equal to the fraction for all values of x. 


859. If x is numerically less than 1, the series is equal to 
the fraction. In this case we can obtain an approximate 
value for the sum of the series by taking the sum of a 
number of terms; the greater the number of terms taken, 
the nearer will this approximate sum approach the value of 
the fraction. The approximate sum will never be exactly 
equal to the fraction, however great the number of terms 
taken ; but by taking enough terms, it can be made to differ 
from the fraction as little as we please. 


Thus, if os, the value of the fraction is 2, and the 
series 1s 


‘Bes clea 
Boies tagtg i au 


The sum of four terms of this series is 14; the sum of 
five terms, 118; the sum of six terms, 134; and so on. 
The successive approximate sums approach, but never 
reach, the finite value 2. 


360, An infinite series is said to be convergent when the 
sum of the terms, as the number of terms is indefinitely in- 
creased, approaches some fixed finite value ; this ae value 
is called the sum of the series. 


324 SCHOOL ALGEBRA. 


3861, In the series lt+at+a2e7tegt... suppose 2 
numerically greater than 1.- In this case, the greater the 
number of terms taken, the greater will their sum be; by 
taking enough terms, we can make their sum as large as 
we pleuse. The fraction, on the other hand, has a definite 
value. Hence, when 2 is numerically greater than 1, the 
series is not equal to the fraction. 

Thus, if #=2, the value of the fraction is —1, and the 
series 1s 


ye a a iy, Si <a 


The greater the number of terms taken, the larger the sum. 
Evidently the fraction and the series are not equal. 


362, In the same series suppose 2 = 1. In this case the 
i i -=5 and the series 1+1+1+14-46-.... 
The more terms we take, the greater will the sum of the 
series be, and the sum of the series does not approach a 
fixed firate value. 

If x, however, is not exactly 1, but is a little less than 1, 


fraction is 





the value of the fraction will be very great, and the 








—2 
fraction will be equal to the series. 
Suppose «=—1. In this case the fraction is j = ; =5, 


and the series 1—1+1—1+..... If we take an even 


number of terms, their sum is 0; if an odd number, their 
sum is 1. Hence the fraction is no¢ equal to the series. 


363. A series is said to be divergent when the sum of the 
terms, as the number of terms is indefinitely increased, 
either increases without end, or oscillates in value without 
approaching any fixed finite value. 


PROPERTIES OF SERIES. 325 


No reasoning can be based on a divergent series; hence, 
in using an infinite series it-is necessary to make such 
restrictions as will cause the series to be convergent. Thus, 
we can use the infinite series 1+xut+a?+ 2° +. when, 
and only when, z lies between +1 and —1. 


364, Identical Series. Jf two series, arranged by powers 
of x, are equal for all values of x that make both series con- 
vergent, the corresponding coefficients are equal each to each. 


For, if A + Bat Ce? = Al4 Blat Cla? + eee, 
by transposition, 

A— A'=(B'—B)x#+(C!-C)v’eo- 

Now, by taking 2 sufficiently small, the right side of this 
equation can be made Jess than any assigned value what- 
ever, and therefore less than A—A!, if A— A! have any 
value whatever. Hence, 4— A! cannot have any value. 

Therefore, 

A— A'=0 or A=A'. 

Hence, Bu + Cz? + D2? + ++ = Bla t+ Ox? + Dia +o 
or (B— B)2=(C'—C)2’?+(D'—D)e2+.---; 
by dividing by 2, 

B-B=(C!'—C)a+(D'—D)x2 po; 
and, by the same proof as for A — A’, 
B— B'=0 o B= SB. 
In like manner, 
Oat A=") and s0-0n. 
Hence, the equation 
At Bat Ce? 4 Alt Bloat Cla? fee, 
if true for all finite values of z, is an identical equation ; 
that is, the coefficients of like powers of x are the same. 


326 SCHOOL ALGEBRA. 


365. Indeterminate Coefficients. 


Ex. Expand pteis 8 Sain ascending powers of zx. 
lt@et+2 
Assume aa aaa =A+ Be + Cr? + Da? 
1+2 + 2 


then, by clearing of fractions, 
24+3¢0=A+4 Brt+ Ce? + Da? + --- 
+ Ax + Bau? + C8 4+ 
+ Aa? + Bad + one 
0 24+30a=A+(B+A)e4+(C+B+A)2?+(D+ C+B) a3 + 
By 3364, A=2, B+A=3, 0+B+A=0, D+C+B=0; 


whence B=1, C=-—38, D=2; and so on. 
by (eee sokcle = 240 —3a2 4 2Qa8 4 
l+a2+2? 


The series is of course equal to the fraction for only such values 
of x as make the series convergent. 


Norr. In employing the method of Indeterminate Coefficients, 
the form of the given expression must determine what powers of the 
variable « must be assumed. It is necessary and sufficient that the 
assumed equation, when simplified, shall have in the right member 
all the powers of x that are found in the left member. 

If any powers of « occur in the right member that are not in the 
left member, the coefficients of these powers in the right member will 

- vanish, so that in this case the method still applies; but if any powers 
of « occur in the left member that are not in the right member, then 
the coefficients of these powers of « must be put equal to 0 in equating 
the coefficients of like powers of x; and this leads to absurd results. 
Thus, if it were assumed that 


243m 


Taeae 5 = Aa + Bo + CF + sees ; 
av 4 67 


there would be in the simplified equation no term on the right cor- 
responding to 2 on the left; so that, in equating the coefficients of 
like powers of x, 2, which is 2x°, would have to be put equal to 02°; 
that is, 2=0, an absurdity. 


PROPERTIES OF SERIES. 827 


Exercise 1410. 


Expand to four terms: 








ie ui ae 4, EAD: 7. eae 
1+ 22 lte«zt+2 l—a2—-2 
2. 2 ' 5. eS Ea, g. Aes atti 
2—32 lta—2 ltete 
3. pe Pe ct sf Laie 
2+32 1—227+32 l—zxz-—62 
Expand to five terms: 

10. ! f ; 12. We ee FN 14, Co Pane 
2+ 24 L+32—2 «(a2 —1)? 
11. 2-2 13. vat) 15, eidinep etd. 

3+ 2 x(a” — 2) (~— D(2+1) 


366, Partial Fractions. To resolve a fraction into partial 
fractions is to express it as the sum of a number of frac- 
tions of which the respective denominators are the factors 
of the denominator of the given fraction. This process is 
the reverse of the process of adding fractions which have 
different denominators. 

Resolution into partial fractions may be easily accom- 
plished by the use of indeterminate coefficients and the 
theorem of § 364. 

In decomposing a given fraction into its simplest partial 
fractions, it is important to determine what form the assumed 
fractions must have. 

Since the given fraction is the swm of the required par- 
tial fractions, each assumed denominator must be a factor 
of the given denominator; moreover, all the factors of the 
given denominator must be taken as denominators of the 
assumed fractions. 


328 SCHOOL ALGEBRA. 


Since the required partial fractions are to be in their 
simplest form incapable of further decomposition, the nu- 
merator of each required fraction must be assumed with 
reference to this condition. Thus, if the denominator is 
a” or (c +a)", the assumed fraction must be of the form 
A or i ES - for if it had the form Ac+ B or Art b B 
a (ea) Lt (z+ a)" 
it could be decomposed into two fractions, and the partial 
fractions would not be in the simplest form possible. 

When all the monomial factors, and all the binomial 
factors, of the form x+:a, have been removed from the 
denominator of the given expression, there may remain 
quadratic factors which cannot be further resolved; and 
the numerators corresponding to these quadratic factors 
may each contain the first power of 2, so that the assumed 
Az+B 

or the 


fractions must have either the form ——————., 
vraxrt+oéb 


Aba Bi 
x’? + 6b 


form 


“__ into partial fractions. 





(1) Resolve 3 


Since 2° +1=(e%+1)(2?—a+1), the denominators will be #+1 
and #—a+1. 





then 3 = A(x? —x +1) + (Be + C)(@ +1) 
=(A+ B)x?+(B+C—A)x+(A+C); 
whence, 3=A+C B+C-—A=0, A+B=0, 
and A=1, B=-1, C=2. 
Therefore, oe ee 


PROPERTIES OF SERIES. 829 


ear Seer Dears ; 
(2) Resolve etc into partial fractions. 
x(x 
The denominators may be 2, x7, «+1, (x +1). 
4e3—a?—3x4—2 A,B C D 
A ‘i Jo: a ial nd ios 
ae oo e+ 1) So ci. wid) 
. 48—2?—-3e—2= Ax(a+1)?4+ B(x+1)? + Ce? (x+1) + Da? 
=(A+C)a®+(2A+B+4+C+D)x?+(A+2B)c+B; 
whence, A+ C=4, 


2A+B+4+0+D=—1 





, 


or B= —2, A= ’ C= 3, D=-—4, 
Therefore, 4a° —2?—32r—2 = i =u 2 3 4 


x? (a# + 1)? Cee +1 (e+1)? 





Exercise 111. 


Resolve into partial fractions : 
Ge Ly Ba 4° 
(7 +4) (#—5) x(a +5) 
Tx—l é (2—o2 
(1—2z)(1— 32) (w—1)? (a+ 2) 
3 5a—l1 2 raat a ll 
| @2—1}(2—5) trae eal 
4 eee 10 Soin ob a ekcagys 
(7 = 5) (z-+ 2) (62-+-1)(¢ —1) 
< te 11. Soin Ect: aessaa 
z?—l (x? + 1) (@ + 2) 
xv—ax—3 v—at+tl 


eed): ar ENS RYN BG 


I. 6 





CHAPTER XXIV. 
BINOMIAL THEOREM. 


367. Binomial Theorem, Positive Integral Exponent. By suc- 
cessive multiplication we obtain the following identities : 

(a+ bP =a + 2ab+ 0’; 

(a+ 6% =a’ + 30° + 8ab’+ 0; 

(a+ b6)*=a*+ 40°) + 607)’ + 4ab* + 0. 

The expressions on the right may be written in a form 
better adapted to show the law of their formation : 


(a+ bf =a + 2ab +2 5h 2 


is 2 3° 2 }? a tae 2. 
(a+ b)°: i 2 ark rary ao : | 
AB ay, 4:3°2 95 | 4:8-2°1 
6y= 4 4 3h 252 53 £ 
at ON Sat eat 3 8. 


Norr. The dot between the Arabic figures means the same as the - 
sign X. 


368, Let represent the exponent of (a+ 6) in any one 
of these identities; then, in the expressions on the right, 
we observe that the following laws hold true: 

I. The number of terms is n+ 1. 

II. The first term is a", and the exponent of a is one 
less in each succeeding term. 

The first power of 5 occurs in the second term, the 
second power in the third term, and the exponent of 6 is 
one greater in each succeeding term. 

The sum of the exponents of a and 6 in any term is n. 


BINOMIAL THEOREM. ook 


III. The coefficient of the first term is 1; of the second 


term, 2; of the third term, ed and so on. 


369. Consider the coefficient of any term; the number 
of factors in the numerator is the same as the number of 
factors in the denominator, and the number of factors in 

each is the same as the exponent of 5 in that term; this 
: exponent is one less than the number of the term. 


370, Proof of the Theorem. ‘That the laws of § 868 hold 
true when the exponent is any positive integer, is shown 
as follows: 

We know that the laws hold for the fourth power ; 
suppose, for the moment, that they hold for the sth power. 

We shall then have 


(a+ 6)* = a* + ka 5 SE) 1) a*-?62 


Hk —1)e—2), enpp cn) 
as Le (1) 


Multiply both members of (1) by a+; the result is 
(apo =ah + (b+ 1a EEE wy 


-@enee Seite Node esene (2) 


In (1) put &+1 for &; this gives 
(a+ by alt + (e+ ats + SEDO EIAD ry: 


Jeet a Dice Sy hh 
=a! + (k+1)a% + aan1, sry o 
ree ena oe ie (3) 


Equation (3) is seen to be the same as equation (2). 


oe 


Fs De SCHOOL ALGEBRA. 


Hence (1) holds when we put &+1 for &; that is, if the 
laws of § 868 hold for the th power, they must hold for 
the (4 + 1)th power. 

But the laws hold for the fourth power; therefore they 
must hold for the fifth power. 

Holding for the fifth power, they must hold for the sixth 
power ; and so on for any positive integral power. 

Therefore they must hold for the nth power, if n is a 
positive integer; and we have 


(a+ b)"=a"+ na" 1h corre 1) a5? 


2 


n(n—1)(n— 


2) qr 3] eesse 


Notr. The above proof is an example of a proof by mathematical 
unduction. 


371, This formula is known as the binomial theorem. 

The expression on the right is known as the expansion of 
(a+6)"; this expansion is a finite series when n is a positive 
integer. ‘That the series is finite may be seen as follows: 

In writing out the successive coefficients we shall finally 
arrive at a coefficient which contains the factor nm —; the 
corresponding term will vanish. The coefficients of all the 
succeeding terms likewise contain the factor »—vn, and 
therefore all these terms will vanish. 


3872, If a and 6 are interchanged, the identity (A) may 
be written 


(a 5 6)" = — (b + a)" = == pt + nb"1a, + —A—— hs “73 — b"-2q? 


angi es ) gn- ds sd oe 


BINOMIAL THEOREM. boo 


This last expansion is the expansion of (A) written in 
reverse order. Comparing the two expansions, we see 
that: the coefficient of the last term is the same as the co- 
efficient of the first term; the coefficient of the last term 
but one is the same as the coefficient of the first term but 
one; and so on. ai 

In general, the coefficient of the rth term from the end 
is the same as the coefficient of the rth term from the 
beginning. In writing out an expansion by the binomial 
theorem, after arriving at the middle term, we can shorten 
the work by observing that the remaining coefficients are 
those already found, taken in reverse order. 


378, If 5 is negative, the terms which involve even 
powers of 6 will be positive, and those which involve odd 
powers of 5 negative. Hence, 


(a — b)" = a" — na" 6 + n “ S 1) qh? 


n(n —1)(n — 2) n—3)3 
Saf hal Se (B) 


Also, putting 1 for a and x for 6, in (A) and (B), 


(1payt= 1 n+ 22a 


Asa LOCA) Pay | 
sins anar Gin tne ©) 


(1—ayt=1— ne - 2D) 


aura (7 L i igo 2 
ieoad ea (D) 


334 SCHOOL ALGEBRA. 


374, Examples: 
(1) Expand (1+ 22). 
In (GO) put 2a for # and 5 for n. The result is 


544.0, 5:4°3 
14228 =1+45(2 oe ae 
(1 42a) 1 4520) +55 48+ 
eae ,5:4°3:2-1aoig 
1°2°3°4 1°2°3°4°5 


=1+4102 4+ 402? + 802° + 80a* + 322°. 


(2) Expand to three terms (; =a 
z 


Put a for = =, and b for 2 ; then, by (B), 


(a — i= a& — 60° + 15.a*b? + «.... 
Replacing a and bd by their values, 


per ah as Be (3) (72) cee 3) 2a7\7 
x 3 x x 3 x 3 


375, Any Required Term. From (A) it is evident (§ 372) 
that the (7+ 1)th term in the expansion of (a+ 60)” is 


n(n —1)(n—2)..... to r factors an" 
[St 233 ear 


Nore. In finding the coefficient of the (r + 1)th term, write down 
“the series of factors 1x 2x 3.....r for the denominator of the co- 
efficient, then write over this series the factors n(n — I}(n — 2), etc., 
writing just as many factors in the numerator as there are in the 
denominator. 


The (7-+1)th term in the expansion of (a—6)” is the 
same as the above if 7 is even, and the negative of the 
above if 7 is odd. 


BINOMIAL THEOREM. 335 


_ Ex. Find the eighth term of (4 ~ ae 


Here a=4, bee n=10, r=—7. 
The term required is eee ee oe 2a Peal 
Pees 24007 BF 
which reduces to — 60214, 


376. A trinomial may be expanded by the binomial 
theorem as follows: 


Expand (1+ 22—27)*. 
Put 2¢—27=2: 
then (1+ 2%=1+4+32+4 32? 4 23, 
Replace z with 2x — a”. 
(1+ 2¢—27)8 =1 +4 3(2e —a?) + 3(2¢ — 2") + (24 — 2?)8 
=1+ 62+ 9a? —42° —9at + 62° — 26. 


Exercise 112. 








Expand : 
1. (a+5)'. 6. (a+ 6)". 11. (m-* 4 nyt, 
pea) 7. (m+n) a9. (ant aly, 
. (8a —Zy). 1 
3 ( x uh 8. we 6°)", 13. (Qa? + 2). 
Ti a 1 2 1 2 
. & = “) 9. (a? + 05)’. 14. (a? —c)*, 
5. (4+ 3y). 10. (at -63) 158 (2a"— iva). 
eorvo 5 ava, VobN' 
Ela 3s) 4S oa coe): 
2a ay doula 
la ars 13V/ ) 3 5 —_— — je 
: (Gta : 2 (rs 9 :) 


18. (22°y— yVy)t. 21. (2ab7 — bat)’. 


336 SCHOOL ALGEBRA. 


Glee tae a \-\On ie dees 
22. onan 23. oleae 
(Va) Naa) 
24. Find the fourth term of (22% — 3y)'. 
25. Find the ninety-seventh term of (2a —b)™. 


Note. As the expansion has 101 terms, the ninety-seventh term 
from the beginning is the fifth term from the end. 


26. Find the eighth term of (82 —y)". 

27. Find the tenth term of (2a? —4a)”. 

28. Find the fifth term of (a—2V)*. 

29. Find the eleventh term of (2— a)”. 

30. Find the fifteenth term of (x + y)”. 

31. Find the fourth term of (8 — 22)’. 

32. Find the twelfth term of (a?— avV 2)". 
33. Find the seventh term of (zy? — 1)*. 

34. Find the fifth term of (4a —b-Vb)". 

35. Find the fourth term of (Va — V2)”. 
36. Find the third term of (Va—-V~— 6)". 
37. Find the sixth term of (Wa? — V—1). 
38. Find the eighth term of (V2a+ V32)™. 
39. Find the ninth term of (e«V—1+yV—1)®. 





3 8 b-? 31 
40. Find the fifth term of (a ie ) : 
Vai 


41. Find the seventh term of (x+ 271)”. 


BINOMIAL THEOREM. B01 


377. Binomial Theorem, Any Exponent. We have seen 
(§ 370) that when » is a positive integer we have the 
identity 

(1-+2)"=1+ ne+ —~—__+ 


n ore 


n(n—1)(n— 2) a 


ae 
ae 1:2°3 


We proceed to the case of fractional and negative expo- 
nents. 


I. Suppose 7 is a positive fraction, 2. We may assume 
that 
(1+ 2%)? = (A+ But Ce? + D2? + oo), (1) 
provided x be so taken that the series 
A+ But Ce? + Da? + 
is convergent (§ 360). 
That this assumption is allowable may be seen as follows: 
Expand both members of (1). We obtain 


—] aah =e 
Sa a 


eeeee 


and A’+qA*'Ba+ ee ») (At? B+ gAtC) ao 


In the first / coefficients of the second series there enter 
only the first & of the coefficients A, B, C, D, ..... If, then, 
we equate the coefficients of corresponding terms in the 
two series (§ 364) as far as the Ath term, we shall have just 
k equations to find k unknown numbers A, B, C, D, ..... 
Hence the assumption made in (1) is allowable. 

Comparing the two first terms and the two second terms, 
we obtain 


At=1, «. A=1: 


= Pp 
ee = DOL. Gis ==), 0)": Bees 


338 SCHOOL ALGEBRA. 


Extracting the gth root of both members of (1), we have 


(1t2y=14+5 a+ Co + Dat + cae (2) 


where x is to be so taken that the series on the right is 
convergent. 


II. Suppose » is a negative number, integral or frac- 
tional, Let »——~m, so that m is positive; then 


1 
1+2)=(1+2)”" = ———. 
( ) ( ) (1 + x)” 
From (2), whether m is integral or fractional, we may 
assume 


1. er 
(lta) 1lt-mzteftdeé po. 


By actual division this gives an equation in the form 


(1+ a)"=1—-— me + Ce? + Da? + + (3) 


378, It appears from (2) and (3) that whether 7 be inte- 


gral or fractional, positive or negative, we may assume 
(1+ a)*=1+ not CP? + Da? +o, 
provided the series on the right 1s convergent. 
Squaring both members, 
(1+ 2¢4+2?)"=1+42ne+2C?'+2 Da? +... (1) 
+ ng? +2nCr’. 
Also, since 
(1+ y)”=1l+nyt CY + DY toe, 
we have, putting 27-+-2’ for y, 
(1+22+2?)"=1+n(22+2*)4+C(22+2°) 
+ D(20 +2)... 
=14+2nr+n2z? +4C7'*+.-.. 
+4 Cv’? + 8 Dz’. (2) 


BINOMIAL THEOREM. 309 
Comparing corresponding coefficients in (1) and (2), 

n+4C=2C+ vn’, 

40+ 8D=2D+2nC 

* 2C=n’?—n, and ga nn= 1) 
122 
(ne 26 ma pet m2). 
Les 
and so on. 


Hence, whether be integral or fractional, positive or 


provided, always, x be so taken that the series on the right 
is convergent. 

The series obtained will be an infinite series unless ” is a 
positive integer (§ 371). 


379, If «x is negative, 


. a 1) her Che Ue Be en oe 
(1—2)"=1—nz+—-~—__+ 19 a 19-3 ge 
Also, if x< a, 


(a+z)"=a ( +2) 


=a” + na" "ze 4 oD arta? oe : 
ioe > a, 


(a-+a)'= (ear ar(1 42) 
SUA Pe aren Bake ) 


oh egy” - 


340 SCHOOL ALGEBRA. 


380, Examples. 


(1) Expand (1 + 28, 
(l+a)@=1+4 o+fG— Do, iG—DG—2) G2) 58 4 


1*2:3 
2 2°5 
= (be eae 
tS sane hi 
The above equation is only true for those values of « which make 
the series convergent. 


1 
PAVE ean tis oe ee 
(2) Exp Fj 


aay i 








1 











Vlas 
law 5 1. 5e a] 
a Pee ae — 4 4 yt ee ee eee 
(—g)et+ Ort + ae x + 
pe 1°59 
aay ie 2 Bot cage 
+4 aa Saletan at 


if x is so taken that the series is convergent. 
A root may often be extracted by means of an expansion. 


(3) Extract the cube root of 344 to six decimal places. 


1 
344 = 343/14 —)=7(14— 
( oe (a cau 


VG ae eee 
343 


t ( 1 
1 4G—)) (ae 
7( 8 alee, e183 a ) 
7(1 + 0,000971815 — 0000000944), 











= 7.006796. 
RA 
(4) Find the eighth term of (2 -- rae me 
4V/x 
Here a=%, Bie oe = n= 1 ye 
4Vu 423 2 





: SAMO Shi . . 11) ee eae Sen 
The term 18 iy Se Ree) MLA) BOSSE We 2 Sie ; 
12:3°4:5°6°7 hy 


or 


BINOMIAL THEOREM. 


Exercise 113. 


Expand to four terms. 


- + x)? 
mkt a). 


. (1+2)7?. 


Peri 25 


. (l—x)- 


16: 
16. 


ks 
18. 
19. 
20. 
21. 


22. 


23. 
24. 


25. 


26. 





6. (L+2)}. ll. (224 8y)?, 
7 (4a) fh, 12. (2x+8y) 4, 
1 
see ae Ea ee LS 
( Vae—ax 
9. (1+ 52)”. 
3 14, Aha 
: 10. (1+52)8. V(a—zx) 
Find the fourth term of (< seh ) 
IV x 
Find the fifth term of ——4—_. 
Via — 22) 


Find the third term of (4 — 7 x)i. 


Find the sixth term of (a — 2ax)s 
Find the fifth term of (1 —22x)~?. 
Find the fifth term of (1 — x)~*. 
Find the seventh term of (1 — x)? 
Find the third term of (1+ 2) an, 


Find the fourth term of (1+ x)~ é 


2 
a 
‘ 


Find the sixth term of (2 -;) 


Find the fifth term of (Qu —8y)~#. 
Find the fourth term of (1 —52)~ 3 





* CHAPTER XXV. 
LOGARITHMS. 


381. If the natural numbers are regarded as powers of 
ten, the exponents of the powers are the Common or Briggs 
Logarithms of the numbers. If A and B& denote natural 
numbers, a and 6 their logarithms, then 10°= A, 10°= B; 
or, written in logarithmic form, log A =a, log B=0. 


382. The logarithm of a product is found by adding the 
logarithms of its factors. 


For AX B= 10 * 10 = ig 
Therefore, log(A x B)=a+b=logA+ log B. 


383, The logarithm of a quotient is found by subtracting 
the logarithm of the divisor from that of the dividend. 


For ay ee TOs: 
Therefore log A _ a—b=logA—log B 
’ 5 B S ols 


384, The logarithm of a power of a number is found by 
multiplying the logarithm of the number by the exponent 
of the power. 


For A" = (10*)" = 10”. : 


Therefore, log A"*=an=nlog A. 


LOGARITHMS. 343 


885. The logarithm of the root of a number is found by 
dividing the logarithm of the number by the index of the 
root. 





For VA = V108 = 10" 
Therefore, log VA= a log A 
n n 


386, The logarithms of 1, 10, 100, etc., and of 0.1, 0.01, 
0.001, etc., are integral numbers. The logarithms of all 
other numbers are fractions. 


mince, 10°= 1; 107° (=7,;) =0.1, 
10F== 10, 107° (=;4,) =0.01, 
10? = 100, 10-° (= apy) = 0.001, 
therefore log 1=0, logO.1 =—l1, 
oo 1 log 0.01 =— 2, 
log 100 = 2, log 0.001 = — 3. 


Also, it is evident that the common logarithms of all 
numbers between 

land 10willbe O- a fraction, 

10 and 100 will be 1-+ a fraction, 

100 and 1000 will be 2-+ a fraction, 

land 0.1 will be —1-a fraction, 

0.1 and 0.01 will be —2-+ a fraction, 

0.01 and 0.001 will be — 8-+a fraction. 


387. If the number is less than 1, the logarithm is nega- 
tive (§ 386), but is written in such a form that the fractional . 
part is always positwe. 


888, Every logarithm, therefore, consists of two parts: a 
positive or negative integral number, which is called the 
characteristic, and a positwe proper fraction, which is called 


344 SCHOOL ALGEBRA. 


the mantissa, Thus, in the logarithm 3.5218, the integral 
number 8 is the characteristic, and the fraction .5218 the 
mantissa. In the logarithm 0.7825 — 2, which is sometimes 
written 2.7825, the integral number — 2 is the character- 
istic, and the fraction .7825 is the mantissa. 


389, If the logarithm has a negative characteristic, it is 
customary to change its form by adding 10, or a multiple 
of 10, to the characteristic, and then indicating the sub- 
traction of the same number from the result. Thus, the 
logarithm 2.7825 is changed to 8.7825 —10 by adding 10 to 
the characteristic and writing —10 after the result. The 
logarithm 18.9278 is changed to 7.9273—20 by adding 20 
to the characteristic and writing — 20 after the result. 


390. The following rules are derived from § 386: 


Rute 1. If the number is greater than 1, make the 
characteristic of the logarithm one wnit less than the num- 
ber of figures on the left of the decimal point. 

Rute 2. If the number is less than 1, make the charac- 
teristic of the logarithm negative, and one unit more than 
the number of zeros between the decimal point and the first 
significant figure of the given number. 

Rute 38. If the characteristic of a given logarithm is 
positive, make the number of figures in the integral part of 
the corresponding number one more than the number of 
units in the characteristic. 

Rute 4. If the characteristic is negatwe, make the num- 
ber of zeros between the decimal point and the first sig- 
nificant figure of the corresponding number one less than 
the number of units in the characteristic. 


Thus, the characteristic of log 7849.27 is 3; the character- 
istic of log 0.037 is —2=8.0000—10. If the characteristic 


LOGARITHMS. 345 


is 4, the corresponding number has five figures in its integral 
part. If the characteristic is — 3, that is, 7.0000 — 10, the 
corresponding fraction has two zeros between the decimal 
point and the first significant figure. 


391, The mantissa of the common logarithm of any inte- 
gral number, or decimal fraction, depends only upon the 
digits of the number, and is unchanged so long as the 
sequence of the digits remains the same. 

For changing the position of the decimal point in a 
number is equivalent to multiplying or dividing the num- 
ber by a power of 10. Its common logarithm, therefore, 
will be increased or diminished by the exponent of that 
power of 10; and since this exponent is mtegral, the man- 
tissa, or decimal part of the logarithm, will be unaffected. 


mame 27196 — 10". 2.7196 = 10°45, 
2G 6 1064. 0.27196 = 10%485-10 
ee ttre. use ne 0027 Ob. == 1 een, 


One advantage of using the number ¢en as the base of a 
system of logarithms consists in the fact that the mantissa 
depends only on the sequence of digits, and the characteristic 
on the position of the decimal pount. 


392. In simplifying the logarithm of a root the equal 
positive and negative numbers to be added to the logarithm 
should be such that the resulting negative number, when 
divided by the index of the root, gives a quotient of — 10. 

Thus, if the log 0.002? = 1 of (7.3010 — 10), the expres- 
sion 4 of (7.3010 — 10) may be put in the form 4 of 
(27.3010 — 30), which is 9.1003 — 10, since the addition of 
20 to the 7, and of — 20 to the — 10, produces no change 
in the value of the logarithm. 


346 SCHOOL ALGEBRA. 


Exercise 114. 


Given: log2=0.38010; log8=0.4771; log 5=0.6990 ; 
log 7 =0,8451. 

Find the common logarithms of the following numbers 
by resolving the numbers into factors, and taking the sum 
of the logarithms of the factors: 


1. log 6. 5. log 25. © 9. log 0.02E "iSie icp 
2. log 15. 6. log 30. 10. log 0.85. 14. log 16, 

3. log 21. 7. log 42. 11. log 0.0035. 15. log 0.056. 
4. log 14. 8. log 420. 12. log 0.004. 16. log 0.63. 


Find the common logarithms of the following : 


iF 
17; 2 <9) 90.58 laa.05t. 26. 77. 29. 5%, 
18: 5X 24.) o3; agate Wied. be 30. 27. 
19.47! © 22) 5ileke o50dos. 28, 310) ates 


3938. The logarithm of the reciprocal of a number is 
called the cologarithm of the number. 
If A denote any number, then 


colog A = log =log1— log A (§ 383) = — log A (§ 386). 


Hence, the cologarithm of a number is equal to the log- 
arithm of the number with the minus sign prefixed, which 
sign affects the entire logarithm, both characteristic and 
mantissa. 

In order to avoid a negative mantissa in the cologarithm, 
it is customary to substitute for — log A its equivalent 
(10 — log A) — 10. 

Hence, the cologarithm of a number is found by subtract- 
ing the logarithm of the number from 10, and then annexing 
— 10 to the remainder. 


LOGARITHMS. 347 


The best way to perform the subtraction is to begin on 
the left and subtract each figure of log A from 9 until we 
reach the last significant figure, which must be subtracted 
from 10. 

If log A is greater in absolute value than 10 and less 
than 20, then in order to avoid a negative mantissa, it is 
necessary to write —log A in the form (20 — log A) — 20. 
So that, in this case, colog A is found by subtracting log A 
from 20, and then annexing — 20 to the remainder. 


(1) Find the cologarithm of 4007. 


10 — 10 
Given: log 4007 = 3.6028 
Therefore, colog 4007 = 6.3972 —10 


(2) Find the cologarithm of 108992000000. 
20 en) 

Given : log 103992000000 = 11.0170 

Therefore, colog 103992000000 = 8.9830 — 20 


If the characteristic of log A is negative, then the subtra- 
hend, —10 or — 20, will vanish in finding the value of 
colog A. 


(3) Find the cologarithm of 0.004007. 


10 —10 
Given: log 0.004007 = 7.6028 — 10 
Therefore, colog 0.004007 = 2.3972 


By using cologarithms the inconvenience of subtracting the 
— logarithm of a divisor is avoided. For dividing by a num- 
ber is equivalent to multiplying by its reciprocal. Hence, 
instead of subtracting the logarithm of a divisor, its colog- 
arithm may be added. 


348 -- - $CHOOL ALGEBRA. 


5 


4) Find the logarithm of 
(4) Find the logarithm o 0.002 





5 
log D008 2 log 5 + colog 0.002. 


~ log 5 = 0.6990 

colog 0.002 = 2.6990 

log quotient = 3.3980 
0.07 


O3 
2 


(5) Find the logarithm of 





log x = log 0.07 + colog 2°. 
log 0.07 = 8.8451 — 10 
colog 2° = (10 — 3 log 2) — 10 = 9.0970 — 10 
log quotient = 7.9421 — 10 


Exercise 115. 
Given: log2=0.38010; log 3=0.4771; log 5 = 0.6990; 
log 7 = 0.8451; log 11 = 1.0414. 
Find the logarithms of the following quotients: 











are Py tk 15, 2:08. 22. oy 
7 ie 44 gi 
3 
2. = 9. a 16. a 23. oS 
: 2 
3. 2. 10. = 123 24, ae 
7 2 
4. x iL 18. = 25. SS 
3 
5. a. 12. 2 19) 26nd 
6. 1s iS; sa 20. = 27. aaa 
: 5 
re z 14. ce 21. o 28. ar 


LOGARITHMS. 349 


394, Tables. A table of fowr-place common logarithms 
is given at the end of this chapter, which contains the com- 
mon logarithms of all numbers under 1000, the decimal point 
and characteristic being omitted. The logarithms of single 
digits, 1, 8, etc., will be found at 10, 80, ete. 

Tables containing logarithms of more places can be pro- 
cured, but this table will serve for many practical uses, and 
will enable the student to use tables of five-place, seven- 
place, and ten-place logarithms, in work that requires 
greater accuracy. 

In working with a four-place table, the numbers corre- 
sponding to the logarithms, that is, the antilogarithms, as 
they are called, may be carried to four significant digits. 


395. To Find the Logarithm of a Number in this Table. 


(1) Suppose it is required to find the logarithm of 65.7. 
In the column headed ‘‘N” look for the first two significant 
figures, and at the top of the table for the third significant 
figure. In the line with 65, and in the column headed 7, 
is seen 8176. To this number prefix the characteristic and 
insert the decimal point. Thus, 


log 65.7 = 1.8176. 


(2) Suppose it is required to find the logarithm of 2034’. 
In the line with 20, and in the column headed 83, is seen 
3075; also in the line with 20, and in the 4 column, is seen 
3096, and the difference between these two is 21. The dif- 
ference between 20300 and 20400 is 100, and the difference 
between 20300 and 20347 is 47. Hence, 45 of 21 = 10, 
nearly, must be added to 3075; that ig, 


log 20347 = 4.3085. 
(3) Suppose it is required to find the logarithm of 


0.0005076. In the line with 50, and in the 7 column, is 
seen 7050; in the 8 column, 7059: the difference is 9. The 


350 SCHOOL ALGEBRA. 


difference between 5070 and 5080 is 10, and the difference 
between 5070 and 5076 is 6. Hence, ;5 of 9=5 must be 
added to 7050; that is, 


log 0.0005076 = 6.7055 — 10. 


396, To Find a Number when its Logarithm is Given. 

(1) Suppose it is required to find the number of which 
the logarithm is 1.9736. 

Look for 9786 in the table. In the column headed ‘“N,” 
and in the line with 9736, is seen 94, and at the head of 
the column in which 9786 stands is seen 1. Therefore, 
write 941, and insert the decimal point as the characteristic 
directs; that is, the number required is 94.1. 


(2) Suppose it is required to find the number of which 
the logarithm is 3.7936. 

Look for 7936 in the table. It cannot be found, but the 
two adjacent mantissas between which it lies are seen to be 
7931 and 7988; their difference is 7, and the difference be- 
tween 7931 and 7936 is 5. Therefore, # of the difference 
between the numbers corresponding to the mantissas, 7931 
and 7938, must be added to the number corresponding to 
the mantissa 7931. 

The number corresponding to the mantissa 7938 is 6220. 

The number corresponding to the mantissa 7931 is 6210. 

The difference between these numbers is 10, 
and 6210 + 3 of 10 = 6217. 

Therefore, the number required is 6217. 


(3) Suppose it is required to find the number of which 
the logarithm is 7.3882 — 10. 

Look for 8882 in the table. It cannot be found, but the 
two adjacent mantissas between which it les are seen to be 
3874 and 8892; the difference between the two mantissas 
is 18, and the difference between 3874 and the given man- 
tissa 3882 is 8. 


LOGARITHMS. ODL 


The number corresponding to the mantissa 3892 is 2450. 
The number corresponding to the mantissa 3874 is 2440. 
The difference between these numbers is 10, 

and 2440 + 58 of 10 = 2444. 
Therefore, the number required is 0.002444. 


Exercise 116. 


Find, from the table, the common logarithms of : 

Peo) ~ “4. ‘7808. Te eNOS: 10. 000.5234. 
2. 201, 5. 4825. re ES 11. 0.01423. 
3. 888. 6. 8109. 9. 0.00789. 12, 0.1987, 


Find antilogarithms to the following common logarithms: 
13. - 4.1482. 15. 2.3177. 17. 9.0380 — 10. 
14. 3.5317. 16. 1.3709. 1B eA ate — 5k). 


397, Examples. 


(1) Find the product of 908.4 x 0.05392 x 2.117. 
log 908.4 = 2.9583 
log 0.05392 = 8.7318 — 10 
log 2.117 = 0.3257 
2.0158 = log 103.7. Ans. 

When any of the factors are negative, find their logarithms with- 
out regard to the signs; write — after the logarithm that corresponds 
to a negative number. If the number of logarithms so marked is 
odd, the product is negative; if even, the product is positive. 


— 8.3709 x 834.637 
7308.946 
log 8.3709=0.9227 — 
log 834.637 = 2.9215 He 
colog 7308.946 = 6.1362 — 10 + 
9.9804 — 10 = log — 0.9558. Ans. 





(2) Find the quotient of 


352, SCHOOL ALGEBRA. 


(3) Find the cube of 0.0497. 
log 0.0497 = 8.6964 — 10 
Multiply by 3, 3 
6.0892 — 10 = log 0.0001228. Ans. 


(4) Find the fourth root of 0.00862. 
log 0.00862 = 7.9355 — 10 
Add 30 — 30, 30, — 30 
Divide by 4, 4.)37.9355 — 40 
9.4839 — 10 = log 0.3047. Ans. 


5} 8.1416 x 4771.21 x 2.71832 
80.108! x 0.48432 x 69.897 


log 3.1416= 0.4971 = 0.4971 

log 4771.21 = 3.6786 = 3.6786 

dlog 2.7183 = 0.4343+2 =0.2172 
4colog 30.103 = 4(8.5214 — 10) = 4.0856 — 10 

dcolog. 0.4343 = 0.3622+2 =0.1811 
4Acolog 69.897 = 4(8.1555 — 10) = 2.6220 — 10 


11.2816 — 20 
30 — 30 


5 ) 41.2816 — 50 


8.2563 — 10 
= log 0.01804. Ans. 


(5) Find the value of 


398. An exponential equation, that is, an equation in which 
the exponent involves the unknown number, is easily solved 


by Logarithms. 


Ex. Find the value of x in 81*= 10. 
8lz = 10, 
. log (81*) = log 10, 
x log 81 = log 10, 
10g 10 _ 1.0000 


= ——_ = 0,524. Ans. 
log 81 1.9085 


LOGARITHMS. 353 


Exercise 117. 


Find by logarithms the following products : 





1. 948.7 x 0.04887. 5. 7564 x (— 0.008764). 
2. 8.409 x 0.008763. 6. 3.764 x (— 0.08349). 
3. 830.7 x 0.0003769. 7. —5.845 x (— 0.00178). 
4. 8.489 x 0.9827. 8. — 8045.7 x 73.84. 
Find by logarithms: 
9, 7065 stiri 45 0.076540) 
5401 83.94 x 0.8395 
See 652 yo, 212 (—6.12) x (— 2008), 
— 0.06875 365 x (— 531) x 2.576 
13. 0.1768, 17, (44)" 21. 2.563%, 
14, 1.211". 18. 906.8%, 22. (83)%*, 
15. 11%. 19. (284)%, 23. (581), 
16. (73)". 20. (748,)™. 24, (929)>. 


25. 1) 0.0075? x 78.84 x 172.44 x 0.00052 
42853 x 54.274 x 0.001 x 86.792 





og. .1]0.03271? x 3.429 x 0.7752 
32.79 x 0.000371! 


‘| 7.126 x V0.1827 x 0.05738 


Qi. N 
\/0.4346 x 17.38 x 0.006372 


Find x from the equations: 
gue.) = 10, RU i herta= 4240 A 32. (0.4)-% = 3. 
29. 4% = 20. 31. (1.3)*=42, 33: (0.9) 7*=2 


004 


10 | 0000 
ll | 0414 
12 | 0792 
13 | 1139 
14 | 1461 


1761 
2041 
2304 
2553 
2788 


3010 
3222 
3424 
3617 
3802 


3979 
4150 
4314 
4472 
4624 


0086 
0492 
0864 
1593 | 


SCHOOL ALGEBRA. 


0128 
0531 
0899 
1229 


1523 | 1553 


1818 
2095 
2355 
2601 
2833 


3054 
3263 
3464 
3655 
3838 


4014 
4183 
4346 
4502 
4654 


1847 
2122 
2380 
2625 
2856 


3075 


0170 
0569 
0934 
1271 
1584 


1875 
2148 
2405 
2648 
2878 


3096 


3284 | 3304 
3483 | 3502 
a 3692 


3856 


4031 
4200 
4362 
4518 
4669 


3874 


4048 
4216 
4378 
4533 
4683 


0212 
0607 
0969 
1303 
1614 


1903 
2175 
2430 
2672 
2900 


3118 
3324 
3522 
3711 
3892 





4065 
4232 
4393 
4548 
4698 





4914 
5051 
5185 
5315 


5441 
5563 
5682 
5798 
5911 





4800 
4942 
5079 
5211 
5340 


5465 
5587 
5705 
5821 
5933 


4814 
4955 
5092 
5224 
5353 


4829 
4969 
5105 
5237 
5366 


4843 
4983 
5119 
5250 
5378 





5478 
5099 
5717 
5832 
5944 


5490 
5611 
5729 
5843 
5955 


5502 
5623 
5740 
5855 
5966 





6021 
6128 
6232 
6335 
6435 


6042 
6149 
6253 
6355 
6454 


6053 
6160 
6263 
6365 
6464 


6064 
6170 
6274 
6375 
6474 


6075 
6180 
6284 
6385 
6484 





6532 
6628 
6721 
6812 
6902 


6990 
7076 
7160 
7243 
7324 | 7332 


7084 
7168 


eas 
feeeee 
i 
auees 
Pale elie 


6951 
6646 
6739 
6830 
6920 


7007 
7093 
7177 
7259 
7340 


AEAee 
z 
faa 
| 


6561 
6656 
6749 
6839 
6928 


7016 
7101 
7185 
7267 
7348 


6571 
6665 
6758 
6848 
6937 


7024 
7110 
7193 
7275 
7356 


6580 
6675 
6767 
6857 
6946 


7033 
7118 
7202 
7284 
7364 





LOGARITHMS. 855 





























CHAPTER XXVI. 


GENERAL REVIEW EXERCISE. 


l'a=6 b=5 cu—4 d= 3. nnd eee 


1. 


2. 


Vb + ac+ Ve — 2ac. 3. VB? tact Ve —2ac. 


a — Vb +a0c 4 e+~vV/di te 
2a—Vb—ae c+ 2d(d*—¢) 


Find the value of 


5. 


10. 


5 


12. 


13. 








ene abe 

—+~ when «= i 

anger when 2 ony, 

Mb) ales een when PR ideal 
a b a a 

os x _a(b—a) 

oo i ak 

maar: , when 2 TOK 

(a+) (6+2)—a(b+¢)+2*, when deeb 





a(l+6)+o62 a 1 yi 
a(l1+6)—be a—2ba renin 5 - 


Add (a —6) 2°+ (b —c) y+ (e—a) 2’, (b—¢) 2+ (e—a)v 
+ (a — b)2, and (ec —a)2*+ (a— b)y¥’+(6—e)2’. 
Add (a+6)a+(b+c)y—(e+a)z, (6+e)z2+(e+a)x 
—(a+b)y, and (a+c)y+ (a+ b)z—(6+0e)2. 

Show that 23+ 7+ 2— 32yz=0, if#+y+z2=0. 
Show that 2*— 87°— 2727— 18 xyz =0, if = 2y+3z2. 


GENERAL REVIEW EXERCISE. ooT 


Simplify by removing parenthesis and collecting terms: 
14. 8a—2(6—c)—[2(a—b)—38(e+a)]—[9e—4(ce—a)]. 
15. 7(2a+ 6) —§19b — [18(e—a) + 12(6 —c)}}. 

16. x—j4y+[8(2—2)—(#4+ 2y)]— (2y+2—22)}. 
17. 14+ 2§4+4—38[4+5—4(2+1])}}. 

18. 10% —§4[52—38(2—1)]—38[42—3(¢#+1)}}. 

19. 32°—{22?—(82—7)—[2 2°—(8 2—2’)|—[5-—(22°—4 2) }}. 
20. (% —2)(x — 8) — (w— 7)(x—1) + (@— 1) (x — 2). 
21. (2+ y) — 22(32-+2y) —(y—2z)(—2+y). 


pomee (2a —(3a—b)) + 3a(2d—8a—") 
Resolve into lowest factors : 
23. (x+y)y— 42’. 27. 92? —8aty?+ yy. 
24, (a? +") —42°y’. 28. OF eae yy 
25. a&—F—e’+2be. 29. 8lat—1. 
26. (#—y?— 2)?— 472. 30. a*— 0b”, 
31. (a —b’+ ¢ —d’) — (2ac — 26d)’. 
32. x’ —192 + 84. 40. 2-7. 
33. 42°?+ Qa — 36. 41. 8+ a°2°, 
34. 2? — 8x+ 15. 42. 2°— a", 
35. 92? — 1502+ 600. 43. 270° — 64. 
36. 562°+ 3xy — 20y’. 44, 2 — 327/. 
37. 1227+ 38744 21. 45. a? — b°. 
38. 33 — 14xz— 402’. 46. 2° + 10247". 
39. 627+52—4. 47. a’ —(b+c)*. 


48. 82°—62xy(2x4+3y) + 27y’. 


358 ’ SCHOOL ALGEBRA. 


Find the H.C. F. of 
49. 6a*—2a°+ 92?+ 92-4, and 92*+ 802? — 9. 
50. 32°—52°+2, and 22°— 527+ 3. 
51. wv — 98a — 808, and 2*® — 212’?4 13812 — 281. 
52. at —2a°+ 42?—62+8, and a*— 22°— 22°+ 62 — 3. 
53. a2 —4¢ — 24-2242, and 2-2 ae 
54. 32° +102?+ Ta —2, and 32°4 1382°4+172+6. 
55. 4a*— 9a?+6x2—1, and 62°—T2’?+1. 
56. 2 +1la—12, and 2° 4+ lla’ 54. 


Find the L.C.M. of 

57. 42°+42—8, and 427+ 24-6. 

58. 2° — 47’, and 2? + 2y— 6y’. 

59. Ta’x(a—x), 2Zlar(a’?— 2’), and 12a27(a+ 2). 

60. 92°—x— 2, and 82° — 102?— Ta —4. 

61. 2?—52+6, 2—424+3, and 2 — 382-4 2. 

62. 2a°-+ 5ary— 5ay+y’, and 2a°— Ta’y-+ day?— y’. 
Simplify : 

sn age Sea ee 


63. inset 
a(x—2). “2(¢+4+1). #—2—2 








BA) 1 si cg OR Sec CA I 
l—x l+e 1l-2# 142 
65 z—l1 2-2) 4 x+5 
(e+2)(a@+5) («+5)@—-1) («—-1]l@+2) 
Fs MMB Wiese cate ee 
axc—a ax+t2e 2wv#+ar—2e 
67 a x—l x—8s 





(w+ Be 1a ee eee (2—«2)(1—2) 


68. 


70. 


(Qs 


T2. 


73. 


80.. 


GENERAL REVIEW EXERCISE. 















































4 12 4 12 
1 ETE oc Ely 
( Petes) Ry) a) 
eee tt ( x eae 
ae Yh Wey y 
feve a-b. T/atd 1 } 
a’— bt 2(ae—0?) 2\e+s? a+b 
1 i, 
dees | fap Ep 
es elias ee 
| es pateee 
a B ew b 
1: 1 75 : i 
Peon 1. a@ x—1-- 
haa 1 
a ‘a 1 sires 
l—a l+a ; 
My 2) 76. 
—#/1—40 —2)] mee a 
i—4[1-+0.—@)] Fe 
SL Na LS 
Solve: 
62+138 92+15 g  2e+1d 
a. 15 ape a, 5 
78 9x +a, Dae be oy 
SS(7-— a). 2 4F a) Ra 
x zt+l «2-8 -w«-9 
1 re == Ps 
: a—-2 x-—1 «2-6 2-T 
can 5a , 0.3 ) 
— apg! BM Se ee 
a | Mii y | 
z., 10 10z , 9 
Ba || Tssnoe ais) 


309 


360 SCHOOL ALGEBRA. 
Aap aeayt Banee a vee 
Got See eee a ee 
me age . | 83. ne 3 
4a, 2y 232 a bie ee 
— Seeahlen Bham eae San eh — — — — 
sg at C zy z 
a ») 
bz ty 225 | 20: 
aero e Lee) Se 


84. 362: 
s5. 32%; 648; 81; (88); (Byy)*;. (1ah)*. 
86. (0.25)?; (0.027)8; 49°; 32%; 81%", 


87. 36-2; 27-3; () ?; (0.16) *; (0.0016) * 


) 
Le Pe ee ars | —3\—-4 
88. Interpretia (savas ea 2 (ae. 
Simplify : 
1 s Rpg 1 1 Bai y RAL oe 
89. a2 a3 CF 08° 998? igh? 5) Ee ee 


ie y a are red 1,24¢e 
90. abc! x a 2b 3c *: abh2e FX ab 2e2d. 


) 





91. (2ab + 2be+ 2ac—a? —B'— 2) + (a? +? + c?). 
92. V12; V8; V50; V16; 4250; Vi; Vi; VS. 
93. 5W— 320; Vaid’; Va; Vax tat; 8154 2°. 
94, 2V/18 — 8V8+ 2-750; V81 + V24— V192. 


95. §V$+ V80—1V20; 8V 35, + 10-V 39 —2V38. 


Rationalize the divisor, and find the value of 





pe es Saal nba sue 100, 2a 
2—V3 V2+V7 Vi+V7 
97. 3 ; 99. V3+v2 101 2V'6 





34-6 VE a2 ' 33 ee 


GENERAL REVIEW EXERCISE. 361 








103. zg+38 2xr-—1 








104. Petes 2 Gs 


pee hb ae 


105. = 
es er a, ab 











goto cated 


106. 
atbz etdz 


107. 











J ba =a+b. 
: 7 a 


109. 
110. az 
111. —— 


112. 














x’ + 10ay = an 
r 9) 


ory —3y' = 


Va 
pps 


Eee a 


ry —xz'*=8 


Form the equations of which the roots are : 


116. a—b, a+. 
117. a— 26, a+ 30. 
118. a+26, 2a-+ 8. 
Solve: | 
122. x*—527+4=—0. 
123. eros B= 
124. 92t'—132°+4=0. 
125. 42*—172?+4=-0. 
126. 22° —5x77+2=0; 


119. 
120. 


121. 


127. 
128. 
129. 
130. 
131. 


oben / at eae 
AK See OF 
Liv 8, V8, 


2a° — 19a? 4+ 24=0, 
z‘—1=0. 

gee Vie), 

x? + 8x3 —9 = 0. 
16z* —17a°>+1=0, 


ee 


362 SCHOOL ALGEBRA. 


132. 2a% —8x2t+1=0. 137. ott be tae. 
133. BVz—5Vz+2=0. 138. 42 $324 27=0, 
134. 6VWx—3V2—45=0. 139. 2”+382"—4=0, 
135. QVet—5V2—74=0. 140. 32° —2ax*—a?=0. 
136. 8V2+4V2—20=0. 141. V2x—~V2r2—2=0. 

142. 83V92°+4V32—39=0. 

143. V3ar+avV3ar —2a°=0. 

144, 3VW2bx — 5bV2br — 28? =0. 

145. V2+4+V8¢+1=V9e4+4. 

146. Vd5¢7+1+4+ 2V42—3 =10V2—2. 

147. V2 +2—8V32—5+ Vb24+1=0 

148. Vll—2+vV8— 22—V214+ 227=0. 

ape: 


3\ 5 
150. (4 (22+-V3 2) (Va2+1+0')* (1+-22—a'—a")". 
151. Expand to four terms 
(1—32) 4, (1— 4a) ?; 1 — gah? (@ 2a 
152. Find the eighty-seventh term of (2a — y)”. 


153. Resolve into partial fractions 





3—2xe 3— 22 . ee 
1—82+22?’ (l—az)(1—82)’ 1-2 
3 — 2x 


154. E d to five t Umeha °F. 
xpand to five ia a ENO eerie 








Ft 
a % 
4“ 
* 
ak 
.s 
t.. 
a Aa 


= —_ a 
ae eel 


5 Seetiel 





UNIVERSITY OF ILLINOIS-URBANA 


12.9W48S1891 C001 
A SCHOOL ALGEBRA BOSTON 


yg 


nad ert 
By W i ry 
6 


eacueas gris 
{ee ween te 


id LF aA 


Ve & 





